New to Forum: Proving f(n+3) -3f(n+2) + 3f(n+1)-f(n) = 0

[Fabio010]

Hi people, as you can see i am new in this forum.

Sorry if i am posting in wrong section.

Im going to try to traduce the problem.

" f(n) is the sum o n terms of a arithmetic progression. "
Show that :

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


I just want to know if

-3f(n+2) = f(-3n -6) ??

 

[dacruick]

Hi Fabio,

couldn't you test this yourself? Why don't you define an arithmetic progression and try it out to see if it works?

EDIT: You could use the simplest arithmetic progression, f(n) = n

 

[Fabio010]

Oh, so easy.

Thanks, i never thought that i could define the arithmetic progression.

 

[HallsofIvy]

Well, what dacruick means is that you can use that particular sequence to "test" your hypothesis that -3f(n+2)= f(-3n- 6). In fact, since f(n) is defined as "the sum of the first n terms of an arithmetic sequence", I can't help but wonder what "f(-3n-6)" means! Of course, that would not prove the theorem that
f(n+3) -3f(n+2) + 3f(n+1)-f(n) = 0 for all n.

for all arithmetic progressions.

Any arithmetic progression is of the form a, a+ d, a+ 2d, ... with nth term a+ d(n-1)

The sum of n terms of that progression is a+ (a+d)+ (a+ 2d)+ ...+ a+d(n-1)= na+ d(1+ 2+...+ (n-1)). It is well known that 1+ 2+ ...+ n-1= n(n-1)/2 so that is f(n)= na+dn(n-1)/2.

Now, calculate f(n+1), f(n+2), f(n+3) and put them into the formula.