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Proof by induction

  1. Oct 30, 2005 #1
    I am supposed to prove a conjecture by induction. I have worked out that the series can be described by:

    a = 2,6,12,20,30
    S = 2+8+20+40....

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

    However, i cannot prove it by induction. It seems like there is smth wrong with the K^3.
  2. jcsd
  3. Oct 30, 2005 #2
    What series? And what precisely is it that you want to prove? I doubt that you actually want to prove that (n^2) + n = (1/3) ((n^3)+3(n^2)+2n), because that's not true (for all n).
  4. Oct 30, 2005 #3


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    Homework Helper

    This should help

    I think Link means [tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)[/tex], which is in fact true.
    The inductive proof goes like this:
    i. [tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)[/tex] , holds for [tex]n=1[/tex];
    ii. Assume [tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)[/tex] holds for some fixed n, so that
    [tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)\Rightarrow \sum_{k=1}^{n} \left(k^2 + k\right) + \left((n+1)^2 + (n+1)\right) = \frac{1}{3}(n^3+3n^2+2n) + \left((n+1)^2 + (n+1)\right)[/tex]
    [tex]\Rightarrow \sum_{k=1}^{n+1} \left(k^2 + k\right) = \frac{1}{3}(n^3+6n^2+11n+6) = \frac{1}{3}\left((n+1)^3+3(n+1)^2+2(n+1)\right) [/tex].
    Therefore, [tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)[/tex] holds for every positive integer n.
    Last edited: Oct 30, 2005
  5. Nov 6, 2005 #4
    thanks! but there is another problem arising from this, regarding the interpretation of a question.

    Based on the results obtained above, how can i calculate 1^2 + 2^2 + 3^2.....+n^2?

    the problem is the word "calculate", do you think they want me to use the same method to derive a general expression for this series, or actually get an numerical value (i dont think its obtainable)? Im the only one in class doin this investigation and I cannot reach my teacher this week so advice on the meaning of this is appreciated.
  6. Nov 6, 2005 #5


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    Well, you now know that
    [tex]\sum_{k=1}^{n} \left(k^2 + k\right)=\sum_{k=1}^nk^2+ \sum_{k=1}^nk = \frac{1}{3}(n^3+3n^2+2n)[/tex]
    If you know a formula for
    [tex]\sum_{k=1}^{n} k[/tex]
    just subtract that from both sides.
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