Let { a n } n>=1 be a sequence, and consider the sequence of first differences B n = a n - a n-1 , n >=2 A). Prove: If {B n }is a constant sequence, then { a n } is a linear sequence (i.e. a n =An + B for some numbers A and B ). b. Use the result from part (a) to prove: If the sequence of second differences of a given sequence { a n } n>=1 is a constant sequence, then { a n } is a quadratic sequence (i.e. a n = An + Bn +C for some numbers A, B,C ). Solution a)Let Bn=c (a constant) so a(n)-a(n-1)=c =>a(n)=c+a(n-1) For n=2, a(2)=c+a(1) a(3)=c+a(2)=c+c+a(1)=2c+a(1)=(3-1)c+a(1) a(4)=c+a(3)=c+2c+a(1)=3c+a(1)=(4-1)c+a(1) .......................... continuing this way we have a(n)=(n-1)c+a(1)=nc+a(1)-c if c=A and a(1)-c=B then a(n)=An+B which is a linear sequence. (proved) .

1. Let { a n } n>=1 be a sequence, and consider the sequence of first differences
B
n = a n - a n-1 , n >=2
A). Prove: If {B n }is a constant sequence, then { a n } is a linear sequence (i.e.
a n =An + B for some numbers A and B ).
b. Use the result from part (a) to prove: If the sequence of second differences of a
given sequence { a n } n>=1 is a constant sequence, then { a n } is a quadratic
sequence (i.e. a n = An + Bn +C for some numbers A, B,C ).
Solution
a)Let Bn=c (a constant) so a(n)-a(n-1)=c =>a(n)=c+a(n-1) For n=2, a(2)=c+a(1)
a(3)=c+a(2)=c+c+a(1)=2c+a(1)=(3-1)c+a(1) a(4)=c+a(3)=c+2c+a(1)=3c+a(1)=(4-1)c+a(1)
.......................... continuing this way we have a(n)=(n-1)c+a(1)=nc+a(1)-c if c=A and a(1)-c=B
then a(n)=An+B which is a linear sequence. (proved)