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Thread: Induction proof of summation

  1. #1
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    Induction proof of summation

    I want to prove $\displaystyle \sum r^3 = \frac{1}{4}n^2(n + 1)^2$ by induction for all positive integers n but I'm having a bit of trouble after a while:

    for n = 1: $\displaystyle 1^3$ = $\displaystyle 1$ so the statement's true for n = 1.

    assume for n = k: $\displaystyle k^3$ = $\displaystyle \frac{1}{4}k^2(k + 1)^2$

    and for n = k = 1:

    $\displaystyle \displaystyle{\frac{1}{4}(k + 1)^2(k + 2)^2 + (k + 1)^3}$

    $\displaystyle \displaystyle{\frac{1}{4}((k + 1)^2(k + 2)^2 + 4(k + 1)^3)}$

    I'm not sure how to make progress from here. Is there anyone who can help out? Thanks
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  2. #2
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    You need to show:
    $\displaystyle

    \frac{1}{4}(k)^2(k + 1)^2 + (k + 1)^3 =
    \frac{1}{4}(k+1)^2(k + 2)^2
    $

    So the LHS is your original sum and the next cubic term, the RHS is
    the same form as your expression, but for k+1
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  3. #3
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    Hello db5vry
    Quote Originally Posted by db5vry View Post
    I want to prove $\displaystyle \sum r^3 = \frac{1}{4}n^2(n + 1)^2$ by induction for all positive integers n but I'm having a bit of trouble after a while:

    for n = 1: $\displaystyle 1^3$ = $\displaystyle 1$ so the statement's true for n = 1.

    assume for n = k: $\displaystyle k^3$ = $\displaystyle \frac{1}{4}k^2(k + 1)^2$

    and for n = k = 1:

    $\displaystyle \displaystyle{\frac{1}{4}(k + 1)^2(k + 2)^2 + (k + 1)^3}$

    $\displaystyle \displaystyle{\frac{1}{4}((k + 1)^2(k + 2)^2 + 4(k + 1)^3)}$

    I'm not sure how to make progress from here. Is there anyone who can help out? Thanks
    Just be a bit more careful with the notation, and it will help you see what to do next. You should include the limits of the summation, so your induction hypothesis should be:

    Let us assume that, for $\displaystyle n=k$
    $\displaystyle \sum_{r=1}^nr^3 = \tfrac{1}{4}n^2(n + 1)^2$
    Therefore:
    $\displaystyle \sum_{r=1}^kr^3 = \tfrac{1}{4}k^2(k + 1)^2$
    Then:
    $\displaystyle \sum_{r=1}^{k+1}r^3 =\sum_{r=1}^kr^3+(k+1)^3$
    $\displaystyle = \tfrac{1}{4}k^2(k + 1)^2+(k+1)^3$

    $\displaystyle = (\tfrac{1}{4}k^2+[k+1])(k + 1)^2$

    $\displaystyle = \tfrac{1}{4}(k^2+4[k+1])(k + 1)^2$

    $\displaystyle = \tfrac{1}{4}(k+2)^2(k + 1)^2$
    So when $\displaystyle n = k+1,\; \sum_{r=1}^nr^3 = \tfrac{1}{4}n^2(n + 1)^2$ is also true.

    Since you have already shown that this is true for $\displaystyle n = 1$, this is all you need to complete the proof.

    Grandad
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