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Functions, Statistics, and Trigonometry problem

  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

    honestrosewater

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    Can you set up the equation? I get
    [tex](n + 1)^{2} = \frac{[n(n + 1)] + [(n + 1)(n + 2)]}{2}[/tex]
    Can you simplify that?
     
  4. Mar 7, 2005 #3
    Thats what my book has for the first step but im confused about where you got the term (n+2) from putting t(n)=(n(n+1))/2 in.
     
  5. Mar 7, 2005 #4

    honestrosewater

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    t(n) = (n(n + 1))/2
    so t(n + 1) = [(n + 1)(n + 1 + 1)]/2.
     
    Last edited: Mar 7, 2005
  6. Mar 7, 2005 #5
    My book has the next step as t(n) + t(n+1)=(1/2)(n+1)(n+(n+2))

    how did it get there
     
  7. Mar 7, 2005 #6

    honestrosewater

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    I have no idea. Of course, since [a/b = a(1/b)] you can see how they can multiply by 1/2 instead of dividing by 2. I don't know about the rest. The next step after
    [tex](n + 1)^{2} = \frac{[n(n + 1)] + [(n + 1)(n + 2)]}{2}[/tex]
    is
    [tex](n + 1)^{2} = \frac{[n^{2} + n] + [n^{2} + 3n + 2]}{2}[/tex]
    Can you take it from there?
    Edit: Do you know how to simplify (n +1)(n + 2)?
     
    Last edited: Mar 7, 2005
  8. Mar 7, 2005 #7
    Yeah I can, thanks alot honestrosewater
     
  9. Mar 8, 2005 #8

    honestrosewater

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    Just curious- does anyone else see how they got
    t(n) + t(n+1)=(1/2)(n+1)(n+(n+2))?
     
  10. Mar 8, 2005 #9

    dextercioby

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    Beside the very obvious

    [tex] t(n)=\frac{n(n+1)}{2} \Rightarrow t(n+1)=\frac{(n+1)(n+2)}{2} [/tex]

    Therefore the sum is

    [tex] t(n)+t(n+1)=\frac{(n+1)}{2}[n+(n+2)] [/tex]

    Daniel.

    P.S.Is there any other simpler way...?
     
    Last edited: Mar 8, 2005
  11. Mar 8, 2005 #10

    honestrosewater

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    The not-skipping-steps-in-between way of adding them is
    [tex]\frac{[n(n + 1)]}{2} + \frac{[(n + 1)(n + 2)]}{2}[/tex]
    What are the actual steps in getting to
    [tex]\frac{(n+1)}{2}[n+(n+2)][/tex]?
    Or at least some of the steps? I don't see them.
     
  12. Mar 8, 2005 #11

    dextercioby

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    How about factoring
    [tex] \frac{n+1}{2} [/tex] ?

    It's in both terms.

    Daniel.
     
  13. Mar 8, 2005 #12

    honestrosewater

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    :redface: ...... thanks.
     
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