Sum of terms of an arithmetic progression.

Question

I just went through this problems in a book.

Let $f(n)$ be the sum of $n$ terms of an arithmetic progression. Show that

$f(n+3)- 3f(n+2)+3f(n+1)-f(n) = 0$.

Is there any other ways instead of write the specifically form of $f(n),f(n+1),...$ then compute?

Thank you!

Answer 1

Your expression is the third forward difference of $f$. It vanishes because $f$ is at most quadratic. In particular, a degree-$d$ polynomial's forward difference is of degree $d-1$ if $d\ge1$, or vanishing if $d=0$.

Answer 2

It isn't necessary to use the formula that explicitly gives the sum for $ \ f(n) \ \ , $ if that's what you're asking. You can just consider the differences between successive values of $ \ f(n) \ \ : \ $ we can see from $ \ f(1) \ = \ a_1 \ $ and $ \ f(2) \ = \ f(1) + (a_1 + 1·d) \ $ that we can form a general difference $ \ f(k+1) - f(k) \ = \ a_1 + (k·d) \ \ . $

We can then write the given expression as $$ f(n+3) \ - \ 3·f(n+2) \ + \ 3·f(n+1) \ - \ f(n) $$ $$ = \ \ [ \ f(n + 3) \ - \ f(n + 2) \ ] \ - \ 2·[ \ f(n + 2) \ - \ f(n + 1) \ ] \ + \ [ \ f(n + 1) \ - \ f(n) \ ] $$ $$ = \ \ [ \ a_1 + (n+2)·d \ ] \ - \ 2·[ \ a_1 + (n+1)·d \ ] \ + \ [ \ a_1 + n·d \ ] $$ $$ = \ \ (n + 2 - 2n - 2 + n)·d \ \ = \ \ 0 \ \ . $$

This may provide some idea of how to go about proving the generalization that J.G. describes.