SlideShare uma empresa Scribd logo
1 de 53
Baixar para ler offline
Mathematical Induction
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
Test: n = 1
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1
Test: n = 1       L.H .S  2
                          1
                         1
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                             1
Test: n = 1       L.H .S  2                R.H .S  2 
                          1                              1
                         1                        1
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                                  1
Test: n = 1       L.H .S  2                     R.H .S  2 
                          1                                   1
                         1                             1
                                L.H .S  R.H .S
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                                  1
Test: n = 1       L.H .S  2                     R.H .S  2 
                          1                                   1
                         1                             1
                                L.H .S  R.H .S

                 1 1       1       1
A n  k  1       2  2  2 
                 22 3     k        k
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                                  1
Test: n = 1       L.H .S  2                     R.H .S  2 
                          1                                   1
                         1                             1
                                L.H .S  R.H .S

                 1 1       1       1
A n  k  1       2  2  2 
                 22 3     k        k
                  1 1            1            1
P n  k  1 1  2  2              2
                 2 3          k  12
                                            k 1
Proof:
     1 1          1          1 1       1   1
1      2            1 2  2  2 
     22 3      k  12     2 3       k k  12
Proof:
     1 1          1          1 1         1   1
1      2            1 2  2  2 
     22 3      k  12     2 3         k k  12
                             1    1
                         2 
                             k k  12
Proof:
     1 1          1          1 1          1   1
1      2            1 2  2  2 
     22 3      k  12      2 3         k k  12
                             1       1
                         2 
                             k k  12
                             k  1  k
                                    2
                         2
                              k k  1
                                       2
Proof:
     1 1          1          1 1          1    1
1      2            1 2  2  2 
     22 3      k  12      2 3          k k  12
                             1        1
                         2 
                             k k  12
                             k  1  k
                                     2
                         2
                              k k  1
                                        2

                              k 2  k 1
                         2
                              k k  1
                                        2
Proof:
     1 1          1          1 1            1      1
1      2            1 2  2  2 
     22 3      k  12      2 3            k k  12
                             1        1
                         2 
                             k k  12
                             k  1  k
                                     2
                         2
                               k k  1
                                        2

                              k 2  k 1
                         2
                              k k  1
                                        2


                                k2  k       1
                         2              
                              k k  1 k k  1
                                       2         2
Proof:
     1 1          1          1 1             1     1
1      2            1 2  2  2 
     22 3      k  12      2 3             k k  12
                             1        1
                         2 
                             k k  12
                             k  1  k
                                     2
                         2
                               k k  1
                                         2

                              k 2  k 1
                         2
                              k k  1
                                         2


                                k2  k        1
                         2               
                              k k  1 k k  1
                                        2        2

                              k k  1
                         2
                              k k  1
                                       2
Proof:
     1 1          1           1 1             1     1
1      2            1 2  2  2 
     22 3      k  12      2 3              k k  12
                              1        1
                         2 
                              k k  12
                              k  1  k
                                      2
                         2
                                k k  1
                                          2

                               k 2  k 1
                         2
                               k k  1
                                          2


                                 k2  k        1
                         2                
                               k k  1 k k  1
                                         2        2

                               k k  1
                         2
                               k k  1
                                        2

                                    1
                            2
                                  k 1
Proof:
  1 1            1           1 1             1     1
1   2              1 2  2  2 
  22 3        k  12      2 3              k k  12
                             1        1
                        2 
                             k k  12
                             k  1  k
                                     2
                        2
                               k k  1
                                         2

                              k 2  k 1
                         2
                              k k  1
                                         2


                                k2  k        1
                        2                
                              k k  1 k k  1
                                        2        2

                              k k  1
                        2
                              k k  1
                                       2

                                   1
                           2
                                 k 1
     1 1            1              1
1  2  2              2
    2 3          k  12
                                 k 1
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
Test: n = 1
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
Test: n = 1   a1  2  2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
Test: n = 1   a1  2  2
A n  k  a k  2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  a k  2

P   n  k  1 ak 1  2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  a k  2

P   n  k  1 ak 1  2
Proof:
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  a k  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  a k  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
           22
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  a k  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
           22
           4
          2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  a k  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
           22
             4
            2
     ak 1  2
iii  The sequences xn and yn are defined by;
                                           xn  y n          2 xn y n
           x1  5, y1  2         xn1             , yn1 
                                              2              xn  y n
     Prove xn yn  10 for n  1
iii  The sequences xn and yn are defined by;
                                           xn  y n          2 xn y n
              x1  5, y1  2      xn1             , yn1 
                                              2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1
iii  The sequences xn and yn are defined by;
                                           xn  y n          2 xn y n
           x1  5, y1  2         xn1             , yn1 
                                              2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52 
                    10
iii  The sequences xn and yn are defined by;
                                           xn  y n          2 xn y n
           x1  5, y1  2         xn1             , yn1 
                                              2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52 
                    10
A n  k  xk yk  10
iii  The sequences xn and yn are defined by;
                                            xn  y n          2 xn y n
             x1  5, y1  2        xn1             , yn1 
                                               2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52 
                    10
A n  k  xk yk  10

P   n  k  1 xk 1 yk 1  10
iii  The sequences xn and yn are defined by;
                                            xn  y n          2 xn y n
             x1  5, y1  2        xn1             , yn1 
                                               2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52 
                    10
A n  k  xk yk  10

P   n  k  1 xk 1 yk 1  10
Proof:
iii  The sequences xn and yn are defined by;
                                                     xn  y n          2 xn y n
               x1  5, y1  2               xn1             , yn1 
                                                        2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52 
                    10
A n  k  xk yk  10

P     n  k  1 xk 1 yk 1  10
Proof:
                   xk  yk  2 xk yk 
    xk 1 yk 1            x  y 
                                      
                   2  k            k 
iii  The sequences xn and yn are defined by;
                                                     xn  y n          2 xn y n
               x1  5, y1  2               xn1             , yn1 
                                                        2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52 
                    10
A n  k  xk yk  10

P     n  k  1 xk 1 yk 1  10
Proof:
                   xk  yk  2 xk yk 
    xk 1 yk 1            x  y 
                                      
                   2  k            k 

               xk y k
               10
iii  The sequences xn and yn are defined by;
                                                     xn  y n          2 xn y n
               x1  5, y1  2               xn1             , yn1 
                                                        2              xn  y n
     Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52 
                    10
A n  k  xk yk  10

P     n  k  1 xk 1 yk 1  10
Proof:
                   xk  yk  2 xk yk 
    xk 1 yk 1            x  y 
                                      
                   2  k            k 

               xk y k
               10
 xk 1 yk 1  10
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2
  L.H .S  a1
        1
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                          1  5 
  L.H .S  a1                    R.H .S        
                                           2 
        1
                                         1.62
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S
  L.H .S  a2
         1
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S              2
                                       1  5 
  L.H .S  a2                 R.H .S        
                                        2 
         1
                                      2.62
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S              2
                                       1  5 
  L.H .S  a2                 R.H .S        
                                        2 
         1
                                      2.62
                  L.H .S  R.H .S
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                                n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                                  1
                                             1  5 
  L.H .S  a1                     R.H .S           
                                              2 
         1
                                          1.62
                  L.H .S  R.H .S                     2
                                              1  5 
  L.H .S  a2                      R.H .S          
                                               2 
         1
                                           2.62
                   L.H .S  R.H .S
                                           k 1                  k
                                 1  5                1  5 
  A n  k  1 & n  k  ak 1               & ak         
                                  2                    2 
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                                n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                                  1
                                             1  5 
  L.H .S  a1                     R.H .S           
                                              2 
         1
                                          1.62
                  L.H .S  R.H .S                     2
                                              1  5 
  L.H .S  a2                      R.H .S          
                                               2 
         1
                                           2.62
                   L.H .S  R.H .S
                                           k 1                  k
                                 1  5                1  5 
  A n  k  1 & n  k  ak 1               & ak         
                                  2                    2 
                                    k 1
                         1 5 
 P   n  k  1 ak 1  
                             
                         2 
Proof:   ak 1  ak  ak 1
Proof:   ak 1  ak  ak 1
                              k    k 1
               1  5  1  5 
                          
                2   2 
Proof:   ak 1  ak  ak 1
                              k     k 1
               1  5  1       5
                              
                2   2           
                        k 1         1         2
                1  5   1    5    1  5  
                                        
                2   2              2      
Proof:   ak 1  ak  ak 1
                              k   k 1
               1  5  1 5
                        
                2   2     
                        k 1   1         2
                1  5   1 
                            5    1  5  
                                  
                2   2    
                                  2      
                       k 1
               1  5   2       4 
                                 2
                2  1  5 1  5  
Proof:   ak 1  ak  ak 1
                              k                k 1
               1  5  1        5
                               
                2   2            
                        k 1          1         2
                1  5   1     5    1  5  
                                         
                2   2               2      
                           k 1
               1      5  2           4 
                                        2
                2         1  5 1  5  
                           k 1
               1      5  2  2 5  4
                                  2 
                2          1  5  
                              k 1
               1  5               62 5 
                                           2
                2                   1  5  
Proof:   ak 1  ak  ak 1
                              k       k 1
               1  5  1        5
                               
                2   2            
                        k 1          1         2
                1  5   1     5    1  5  
                                         
                2   2               2      
                           k 1
               1      5  2           4 
                                        2
                2         1  5 1  5  
                           k 1
               1      5  2  2 5  4
                                  2 
                2          1  5  
                              k 1
               1  5   6  2 5 
                               
                  2   1  5 2 
                        k 1
               1  5 
                    
                2 
Proof:     ak 1  ak  ak 1
                                k       k 1
                 1  5  1        5
                                 
                  2   2            
                          k 1          1         2
                  1  5   1     5    1  5  
                                           
                  2   2               2      
                             k 1
                 1      5  2           4 
                                          2
                  2         1  5 1  5  
                             k 1
                 1      5  2  2 5  4
                                    2 
                  2          1  5  
                                k 1
                   1  5   6  2 5 
                                   
                      2   1  5 2 
                            k 1
                   1  5 
                        
                    2 
                            k 1
                   1  5 
          ak 1        
                    2 
Proof:           ak 1  ak  ak 1
                                      k       k 1
                       1  5  1        5
                                       
                        2   2            
                                k 1          1         2
                        1  5   1     5    1  5  
                                                 
                        2   2               2      
                                   k 1
                       1      5  2           4 
     Sheets                                     2
                        2         1  5 1  5  
                                   k 1
         +             1      5  2  2 5  4
                                          2 
                        2          1  5  
 Exercise 10E*
                                      k 1
                        1  5   6  2 5 
                                        
                           2   1  5 2 
                                 k 1
                        1  5 
                             
                         2 
                                 k 1
                        1  5 
               ak 1        
                         2 

Mais conteúdo relacionado

Destaque

3 2 absolute value equations-x
3 2 absolute value equations-x3 2 absolute value equations-x
3 2 absolute value equations-xmath123b
 
Plate Tectonics Notes
Plate Tectonics NotesPlate Tectonics Notes
Plate Tectonics Notesduncanpatti
 
Changes In Education
Changes In EducationChanges In Education
Changes In Educationbritte2204
 
Prueba de la teoría de la deriva continental
Prueba de la teoría de la deriva continentalPrueba de la teoría de la deriva continental
Prueba de la teoría de la deriva continentalEdu 648
 
Class at PSTTI on "Math Skills"
Class at PSTTI on "Math Skills"Class at PSTTI on "Math Skills"
Class at PSTTI on "Math Skills"PSTTI
 
Cv312 g formation-db2-for-z-os-new-features-in-version-10-workshop
Cv312 g formation-db2-for-z-os-new-features-in-version-10-workshopCv312 g formation-db2-for-z-os-new-features-in-version-10-workshop
Cv312 g formation-db2-for-z-os-new-features-in-version-10-workshopCERTyou Formation
 
Final semestral alice ciencia
Final semestral alice cienciaFinal semestral alice ciencia
Final semestral alice cienciaSUSY84
 
Recommendation for successful “the saem” facebook page
Recommendation for successful “the saem” facebook pageRecommendation for successful “the saem” facebook page
Recommendation for successful “the saem” facebook pageFloria Hong
 
New features in .NET 4.5, C# and VS2012
New features in .NET 4.5, C# and VS2012New features in .NET 4.5, C# and VS2012
New features in .NET 4.5, C# and VS2012Subodh Pushpak
 
Modulo2.T3.Que necesito para tener un blog
Modulo2.T3.Que necesito para tener un blogModulo2.T3.Que necesito para tener un blog
Modulo2.T3.Que necesito para tener un blogProfesorOnline
 
Analisis laporan keuangan
Analisis laporan keuanganAnalisis laporan keuangan
Analisis laporan keuanganadelaa09
 
LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861
LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861
LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861Marcial Pons Argentina
 

Destaque (19)

3 2 absolute value equations-x
3 2 absolute value equations-x3 2 absolute value equations-x
3 2 absolute value equations-x
 
Plate Tectonics Notes
Plate Tectonics NotesPlate Tectonics Notes
Plate Tectonics Notes
 
Teoria tributaria
Teoria tributariaTeoria tributaria
Teoria tributaria
 
Changes In Education
Changes In EducationChanges In Education
Changes In Education
 
Bio3º
Bio3ºBio3º
Bio3º
 
Recomendaciones valiosas
Recomendaciones valiosasRecomendaciones valiosas
Recomendaciones valiosas
 
Optimistasiempre
OptimistasiempreOptimistasiempre
Optimistasiempre
 
Acta sesion evaluacion_2
Acta sesion evaluacion_2Acta sesion evaluacion_2
Acta sesion evaluacion_2
 
Prueba de la teoría de la deriva continental
Prueba de la teoría de la deriva continentalPrueba de la teoría de la deriva continental
Prueba de la teoría de la deriva continental
 
Class at PSTTI on "Math Skills"
Class at PSTTI on "Math Skills"Class at PSTTI on "Math Skills"
Class at PSTTI on "Math Skills"
 
Cv312 g formation-db2-for-z-os-new-features-in-version-10-workshop
Cv312 g formation-db2-for-z-os-new-features-in-version-10-workshopCv312 g formation-db2-for-z-os-new-features-in-version-10-workshop
Cv312 g formation-db2-for-z-os-new-features-in-version-10-workshop
 
Final semestral alice ciencia
Final semestral alice cienciaFinal semestral alice ciencia
Final semestral alice ciencia
 
Ruta de cuba
Ruta de cubaRuta de cuba
Ruta de cuba
 
Recommendation for successful “the saem” facebook page
Recommendation for successful “the saem” facebook pageRecommendation for successful “the saem” facebook page
Recommendation for successful “the saem” facebook page
 
New features in .NET 4.5, C# and VS2012
New features in .NET 4.5, C# and VS2012New features in .NET 4.5, C# and VS2012
New features in .NET 4.5, C# and VS2012
 
Modulo2.T3.Que necesito para tener un blog
Modulo2.T3.Que necesito para tener un blogModulo2.T3.Que necesito para tener un blog
Modulo2.T3.Que necesito para tener un blog
 
Resume 123
Resume 123Resume 123
Resume 123
 
Analisis laporan keuangan
Analisis laporan keuanganAnalisis laporan keuangan
Analisis laporan keuangan
 
LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861
LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861
LA LUCHA POR LOS DERECHOS.AUTOR: José María Enríquez Sánchez.ISBN: 9788416402861
 

Semelhante a X2 T08 02 induction

X2 T08 02 induction (2011)
X2 T08 02 induction (2011)X2 T08 02 induction (2011)
X2 T08 02 induction (2011)Nigel Simmons
 
X2 t08 02 induction (2012)
X2 t08 02 induction (2012)X2 t08 02 induction (2012)
X2 t08 02 induction (2012)Nigel Simmons
 
11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)Nigel Simmons
 
11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)Nigel Simmons
 
11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)Nigel Simmons
 
X2 t08 02 induction (2013)
X2 t08 02 induction (2013)X2 t08 02 induction (2013)
X2 t08 02 induction (2013)Nigel Simmons
 
11X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 211X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 2Nigel Simmons
 
11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)Nigel Simmons
 
mathematical induction
mathematical inductionmathematical induction
mathematical inductionankush_kumar
 
11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)Nigel Simmons
 
mathematical induction
mathematical inductionmathematical induction
mathematical inductionankush_kumar
 
11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)Nigel Simmons
 
11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)Nigel Simmons
 

Semelhante a X2 T08 02 induction (14)

X2 T08 02 induction (2011)
X2 T08 02 induction (2011)X2 T08 02 induction (2011)
X2 T08 02 induction (2011)
 
X2 t08 02 induction (2012)
X2 t08 02 induction (2012)X2 t08 02 induction (2012)
X2 t08 02 induction (2012)
 
11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)
 
11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)
 
11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)
 
X2 t08 02 induction (2013)
X2 t08 02 induction (2013)X2 t08 02 induction (2013)
X2 t08 02 induction (2013)
 
11X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 211X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 2
 
11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)
 
2.4 edited1
2.4 edited12.4 edited1
2.4 edited1
 
mathematical induction
mathematical inductionmathematical induction
mathematical induction
 
11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)
 
mathematical induction
mathematical inductionmathematical induction
mathematical induction
 
11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)
 
11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)
 

Mais de Nigel Simmons

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATENigel Simmons
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)Nigel Simmons
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)Nigel Simmons
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)Nigel Simmons
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)Nigel Simmons
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)Nigel Simmons
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)Nigel Simmons
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)Nigel Simmons
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)Nigel Simmons
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)Nigel Simmons
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)Nigel Simmons
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)Nigel Simmons
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)Nigel Simmons
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)Nigel Simmons
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)Nigel Simmons
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)Nigel Simmons
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)Nigel Simmons
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)Nigel Simmons
 

Mais de Nigel Simmons (20)

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
 

Último

Introduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxIntroduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxpboyjonauth
 
Paris 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activityParis 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activityGeoBlogs
 
Presiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha electionsPresiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha electionsanshu789521
 
A Critique of the Proposed National Education Policy Reform
A Critique of the Proposed National Education Policy ReformA Critique of the Proposed National Education Policy Reform
A Critique of the Proposed National Education Policy ReformChameera Dedduwage
 
Introduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher EducationIntroduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher Educationpboyjonauth
 
The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13Steve Thomason
 
Q4-W6-Restating Informational Text Grade 3
Q4-W6-Restating Informational Text Grade 3Q4-W6-Restating Informational Text Grade 3
Q4-W6-Restating Informational Text Grade 3JemimahLaneBuaron
 
Separation of Lanthanides/ Lanthanides and Actinides
Separation of Lanthanides/ Lanthanides and ActinidesSeparation of Lanthanides/ Lanthanides and Actinides
Separation of Lanthanides/ Lanthanides and ActinidesFatimaKhan178732
 
mini mental status format.docx
mini    mental       status     format.docxmini    mental       status     format.docx
mini mental status format.docxPoojaSen20
 
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxContemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxRoyAbrique
 
Hybridoma Technology ( Production , Purification , and Application )
Hybridoma Technology  ( Production , Purification , and Application  ) Hybridoma Technology  ( Production , Purification , and Application  )
Hybridoma Technology ( Production , Purification , and Application ) Sakshi Ghasle
 
_Math 4-Q4 Week 5.pptx Steps in Collecting Data
_Math 4-Q4 Week 5.pptx Steps in Collecting Data_Math 4-Q4 Week 5.pptx Steps in Collecting Data
_Math 4-Q4 Week 5.pptx Steps in Collecting DataJhengPantaleon
 
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdfBASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdfSoniaTolstoy
 
Sanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfSanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfsanyamsingh5019
 
Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)eniolaolutunde
 
Incoming and Outgoing Shipments in 1 STEP Using Odoo 17
Incoming and Outgoing Shipments in 1 STEP Using Odoo 17Incoming and Outgoing Shipments in 1 STEP Using Odoo 17
Incoming and Outgoing Shipments in 1 STEP Using Odoo 17Celine George
 
Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111Sapana Sha
 
Crayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon ACrayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon AUnboundStockton
 

Último (20)

Introduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxIntroduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptx
 
Paris 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activityParis 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activity
 
Presiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha electionsPresiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha elections
 
A Critique of the Proposed National Education Policy Reform
A Critique of the Proposed National Education Policy ReformA Critique of the Proposed National Education Policy Reform
A Critique of the Proposed National Education Policy Reform
 
Introduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher EducationIntroduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher Education
 
The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13
 
Q4-W6-Restating Informational Text Grade 3
Q4-W6-Restating Informational Text Grade 3Q4-W6-Restating Informational Text Grade 3
Q4-W6-Restating Informational Text Grade 3
 
Separation of Lanthanides/ Lanthanides and Actinides
Separation of Lanthanides/ Lanthanides and ActinidesSeparation of Lanthanides/ Lanthanides and Actinides
Separation of Lanthanides/ Lanthanides and Actinides
 
mini mental status format.docx
mini    mental       status     format.docxmini    mental       status     format.docx
mini mental status format.docx
 
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxContemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
 
Hybridoma Technology ( Production , Purification , and Application )
Hybridoma Technology  ( Production , Purification , and Application  ) Hybridoma Technology  ( Production , Purification , and Application  )
Hybridoma Technology ( Production , Purification , and Application )
 
Model Call Girl in Bikash Puri Delhi reach out to us at 🔝9953056974🔝
Model Call Girl in Bikash Puri  Delhi reach out to us at 🔝9953056974🔝Model Call Girl in Bikash Puri  Delhi reach out to us at 🔝9953056974🔝
Model Call Girl in Bikash Puri Delhi reach out to us at 🔝9953056974🔝
 
_Math 4-Q4 Week 5.pptx Steps in Collecting Data
_Math 4-Q4 Week 5.pptx Steps in Collecting Data_Math 4-Q4 Week 5.pptx Steps in Collecting Data
_Math 4-Q4 Week 5.pptx Steps in Collecting Data
 
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdfBASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdf
 
Sanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfSanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdf
 
Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)
 
Código Creativo y Arte de Software | Unidad 1
Código Creativo y Arte de Software | Unidad 1Código Creativo y Arte de Software | Unidad 1
Código Creativo y Arte de Software | Unidad 1
 
Incoming and Outgoing Shipments in 1 STEP Using Odoo 17
Incoming and Outgoing Shipments in 1 STEP Using Odoo 17Incoming and Outgoing Shipments in 1 STEP Using Odoo 17
Incoming and Outgoing Shipments in 1 STEP Using Odoo 17
 
Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111
 
Crayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon ACrayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon A
 

X2 T08 02 induction

  • 2. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n
  • 3. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n Test: n = 1
  • 4. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 Test: n = 1 L.H .S  2 1 1
  • 5. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1
  • 6. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1  L.H .S  R.H .S
  • 7. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1  L.H .S  R.H .S 1 1 1 1 A n  k  1   2  2  2  22 3 k k
  • 8. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1  L.H .S  R.H .S 1 1 1 1 A n  k  1   2  2  2  22 3 k k 1 1 1 1 P n  k  1 1  2  2     2 2 3 k  12 k 1
  • 9. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12
  • 10. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12 1 1  2  k k  12
  • 11. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2
  • 12. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2
  • 13. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2
  • 14. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2 k k  1  2 k k  1 2
  • 15. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2 k k  1  2 k k  1 2 1  2 k 1
  • 16. Proof: 1 1 1 1 1 1 1 1  2   1 2  2  2  22 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2 k k  1  2 k k  1 2 1  2 k 1 1 1 1 1 1  2  2     2 2 3 k  12 k 1
  • 17. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1
  • 18. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1
  • 19. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2
  • 20. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  a k  2
  • 21. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  a k  2 P n  k  1 ak 1  2
  • 22. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  a k  2 P n  k  1 ak 1  2 Proof:
  • 23. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  a k  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak
  • 24. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  a k  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak  22
  • 25. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  a k  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak  22  4 2
  • 26. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  a k  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak  22  4 2  ak 1  2
  • 27. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1
  • 28. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1
  • 29. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52   10
  • 30. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52   10 A n  k  xk yk  10
  • 31. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52   10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10
  • 32. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52   10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:
  • 33. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52   10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:  xk  yk  2 xk yk  xk 1 yk 1   x  y     2  k k 
  • 34. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52   10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:  xk  yk  2 xk yk  xk 1 yk 1   x  y     2  k k   xk y k  10
  • 35. iii  The sequences xn and yn are defined by; xn  y n 2 xn y n x1  5, y1  2 xn1  , yn1  2 xn  y n Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52   10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:  xk  yk  2 xk yk  xk 1 yk 1   x  y     2  k k   xk y k  10  xk 1 yk 1  10
  • 36. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2 
  • 37. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2
  • 38. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 L.H .S  a1 1
  • 39. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62
  • 40. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S
  • 41. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S L.H .S  a2 1
  • 42. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62
  • 43. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62  L.H .S  R.H .S
  • 44. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62  L.H .S  R.H .S k 1 k 1  5  1  5  A n  k  1 & n  k  ak 1    & ak     2   2 
  • 45. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62  L.H .S  R.H .S k 1 k 1  5  1  5  A n  k  1 & n  k  ak 1    & ak     2   2  k 1 1 5  P n  k  1 ak 1      2 
  • 46. Proof: ak 1  ak  ak 1
  • 47. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5       2   2 
  • 48. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2   
  • 49. Proof: ak 1  ak  ak 1 k k 1 1  5  1 5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5   2 4      2  2  1  5 1  5  
  • 50. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4      2  2  1  5 1  5   k 1 1  5  2  2 5  4    2   2   1  5   k 1 1  5  62 5     2  2   1  5  
  • 51. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4      2  2  1  5 1  5   k 1 1  5  2  2 5  4    2   2   1  5   k 1 1  5   6  2 5       2   1  5 2  k 1 1  5     2 
  • 52. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4      2  2  1  5 1  5   k 1 1  5  2  2 5  4    2   2   1  5   k 1 1  5   6  2 5       2   1  5 2  k 1 1  5     2  k 1 1  5   ak 1     2 
  • 53. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4  Sheets     2  2  1  5 1  5   k 1 + 1  5  2  2 5  4    2   2   1  5   Exercise 10E* k 1 1  5   6  2 5       2   1  5 2  k 1 1  5     2  k 1 1  5   ak 1     2 