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Math Help - Summation formula

  1. #1
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    Summation formula

    Some help with the following would be appreciated!

    Find (and prove) a formula for Σ from j=1 to k of j^3


    Known propositions:

    Σ from j=1 to k of j = k(k+1) / 2
    Σ from j=1 to k of j^2 = k(k+1)(2k+1) / 6

    Thank you!
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Note that

    \displaystyle\sum_{j=1}^{k+1}j^4 = \sum_{j=0}^{k} (j+1)^4 = \sum_{j=0}^{k}(j^4+4j^3+6j^2+4j+1).

    Hence

    \displaystyle(k+1)^4 = 4\sum_{j=0}^kj^3 + 6\sum_{j=0}^kj^2 + 4\sum_{j=0}^kj + k+1.

    Now use what you know!
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  3. #3
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    Helolo, jstarks44444!

    \displaystyle \text{Find (and prove) a formula for: }\: \sum^n_{k=1}k^3

    \text{Known formulas:}

    . . \displaystyle \sum^n_{k=1}1 \:=\:n \qquad \sum^n_{k=1}k \:=\:\frac{n(n\!+\!1)}{2} \qquad \sum^n_{k=1}k^2 \:=\:\frac{k(k\!+\!1)(2k\!+\!1)}{6}

    We note that: . k^4 - (k-1)^4 \;=\;4k^3 - 6k^2 + 4k - 1


    Let k \:=\:n,\:n\!-\!1,\:n\!-\!2\: \hdots \:3,\:2,\:1


    And we have this "stack" of equations:

    . . \begin{array}{cccccccccccc}<br />
n^4 &-& (n\!-\!1)^4 &=& 4n^3 &-& 6n^2 &+& 4n &-& 1 \\<br />
(n\!-\!1)^4 &-& (n\!-\!2)^4 &=& 4(n\!-\!1)^3 &-& 6(n\!-\!1)^2 &+& 4(n\!-\!1) &-& 1 \\<br />
(n\!-\!2)^4 &-& (n\!-\!3)^4 &=& 4(n\!-\!2)^3 &-& 6(n\!-\!2)^2 &+& 4(n\!-\!2) &-& 1 \\<br />
\vdots && \vdots && \vdots && \vdots && \vdots && \vdots \\<br />
3^4 &-& 2^4 &=& 4\!\cdot\!3^3 &-& 6\!\cdot\!3^2 &+& 4\!\cdot\!3 &-& 1 \\<br />
2^4 &-& 1^4 &=& 4\!\cdot\!2^3 &-& 6\!\cdot\!2^2 &+& 4\!\cdot\!2 &-& 1 \\<br />
1^4 &-& 0^4 &=& 4\!\cdot\!1^3 &-& 6\!\cdot\!1^2 &+& 4\!\cdot\!1 &-& 1 \end{array}


    Add the entire stack of equations and we have:

    . . \displaystyle n^4 \;=\;4\sum^n_{k=1} k^3 - 6\sum^n_{k=1} k^2 + 4\sum^n_{k=1} k - \sum^n_{k=1} 1


    Substitute the known formulas:

    . . \displaystyle n^4 \;=\;4\sum k^3 - 6\!\cdot\!\frac{n(n+1)(2n+1)}{6} + 4\!\cdot\!\frac{n(n+1)}{2} - n

    . . \displaystyle n^4 \;=\;4\sum k^3 - n(n+1)(2n+1) + 2n(n+1) - n


    Then we have:

    . . \displaystyle 4\sum k^3 \;=\;n^4 + n + n(n+1)(2n+1) - 2n(n+1)

    . . . . . . . . =\;n(n^3+1) + n(n+1)(2n+1) - 2n(n+1)

    . . . . . . . . =\;n(n+1)(n^2-n+1) +  n(n+1)(2n+1) - 2n(n+1)

    . . . . . . . . =\; n(n+1)\bigg[(n^2-n+1) + (2n+1) - 2 \bigg]

    . . . . . . . . =\;n(n+1)(n^2+n)

    . . . . . . . . =\;n^2(n+1)^2


    \displaystyle \text{Therefore: }\;\sum^n_{k=1} k^3 \;=\;\frac{n^2(n+1)^2}{4}

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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by jstarks44444 View Post
    Some help with the following would be appreciated!

    Find (and prove) a formula for Σ from j=1 to k of j^3


    Known propositions:

    Σ from j=1 to k of j = k(k+1) / 2
    Σ from j=1 to k of j^2 = k(k+1)(2k+1) / 6

    Thank you!
    Let's write...

    \displaystyle S_{k,n} = 1^{k}+2^{k}+...+n^{k} (1)

    ... where n \in \mathbb{N} and k \in \mathbb{N}. Is...

    \displaystyle \binom{k+1}{1}\ S_{1,n} + \binom{k+1}{2}\ S_{2,n}+ ... + \binom{k+1}{k}\ S_{k,n} = (n+1)^{k+1} - (n+1) (2)

    Now if You know S_{1,n} and S_{2,n} the (2) permits You to find S_{3,n} ... and after that S_{4,n} , S_{5,n} and so one ...

    Kind regards

    \chi \sigma
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