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Basic question.

  1. Oct 24, 2011 #1
    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) ??
  2. jcsd
  3. Oct 24, 2011 #2
    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
  4. Oct 24, 2011 #3
    Oh, so easy.

    Thanks, i never thought that i could define the arithmetic progression.
  5. Oct 24, 2011 #4


    User Avatar
    Science Advisor

    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.
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