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Math Question, need help badly.

  1. Mar 7, 2005 #1
    I've been stuck on this question for awhile.

    Q: Square numbers 1, 4, 9, 16, 25.... are the values of the function s(n)=n^2, when n is a positive integer. The triangular numbers t(n)=(n(n+1))/2 are the numbers t(1)=1, t(2)=3, t(3)=6, t(4)=10.

    Prove: For all positive integers n, s(n+1) = t(n) + t(n+1)

    I've tride alot of things and come to the conclusion that I cant get my answer by using polynomials. I think that if you subsitiute t(n)=(n(n+1))/2 into the equation and simplify to get (n+1)^2 I will be done. My problem is that I'm having troubles doing this. Any sugestions???
     
  2. jcsd
  3. Mar 7, 2005 #2
    im not completely sure what your talking about, but i think your equation is wrong

    [tex]
    ( \frac{1} {2} ( n - 1 ) n ) ^ 2 [/tex] is most similiar to what your talking about, personaly i prefer the: [tex]( \frac{n} {2} ( n + 1 ) ) ^ 2[/tex]

    orr this could also be it: [tex]S_N = \frac{N}{2} ( A_1 + A_N)[/tex]

    and again im only regurgitating equations on you that look like what you could be searching for
     
    Last edited: Mar 7, 2005
  4. Mar 7, 2005 #3
    expand the right hand side and rearrange it into the form of (n+1)^2 = s(n+1)

    note:
    RHS = t(n)+t(n+1) = n(n+1)/2 + (n+1)((n+1)+1)/2
     
  5. Mar 7, 2005 #4
    After changing n to (n+1), you get:

    (n+1)^2 = n(n+1)/2 + (n+1)((n+1)+1)/2)

    = (n+1)^2 = (n^2+n)/2 + (n+1)(n+2)/2

    = (n+1)^2 = (n^2+n)/2 + (n^2+3n+2)/2

    Since they have common denominators, we can add the right side together:

    = (n+1)^2 = (2n^2 + 4n + 2)/2

    = (n+1)^2 = (2n+2)(n+1)/2

    = (n+1)^2 = 2(n+1)(n+1)/2

    The two's cancel out, which gives the needed proof:

    = (n+1)^2 = (n+1)(n+1)
     
  6. Mar 7, 2005 #5
    how did you go from
    = (n+1)^2 = (2n^2 + 4n + 2)/2

    to

    = (n+1)^2 = (2n+2)(n+1)/2
     
  7. Mar 7, 2005 #6
    Thanks for the help all, I got it.
     
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