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Today I saw the following example and I was wondering if it's true for:

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$$\sum_{k=1}^{n} k^{m} = \left ( \sum_{k=1}^{n} k \right )^{m-1}$$

Can someone help me out with it?

Thanks!

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    $\begingroup$ $1^4+2^4 = 17\neq (1+2)^3=27$ $\endgroup$
    – JMoravitz
    Nov 10, 2016 at 3:06
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    $\begingroup$ Because of the degrees of $n$, it can only be a solution for $m+1=2(m-1)$ and therefore only for $m=3$ . $\endgroup$
    – user90369
    Nov 10, 2016 at 11:05
  • $\begingroup$ It seems that you have completely different question in the title and in the body of your question - the tile says (probably?) $\sum_{k=1}^n k^3=(\sum_{k=1}^n k)^2$ and in the body of your question you ask about $\sum_{k=1}^n k^{m}=(\sum_{k=1}^n k)^{m-1}$. $\endgroup$ Nov 11, 2016 at 10:16

2 Answers 2

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It's not. $\sum k^n$ is a polynomial of degree $n+1$. $\left(\sum k\right)^m$ is a polynomial of degree $2m$. They can't match unless $n+1=2m$. (Even if $n+1=2m$ they still might not match.)

You should check out the Wikipedia article on Faulhaber's formula.

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As already answered by MJD, considering $$S_m=\sum_{k=1}^{n} k^{m}\qquad T_m = \left ( \sum_{k=1}^{n} k \right )^{m-1}\qquad \Delta_m =T_m-S_m $$ and using Faulhaber's formulae , you would find $$\Delta_1=1-\frac{1}{2} n (n+1) $$ $$\Delta_2=-\frac{(n-1)n(n+1)}{3} $$ $$\Delta_3=0$$ $$\Delta_4=\frac{(n-1) n (n+1)(15 n^3+21 n^2-4)}{120} $$ and so on.

Edit

As MJD commented, it is probably more interesting to look at $\Phi(m)=S_{2m-1} - T_{m+1}$. Some results are given below $$\Phi(1)=0$$ $$\Phi(2)=0$$ $$\Phi(3)=\frac{1}{24} n^2 (n+1)^2 \left(n^2+n-2\right)$$ $$\Phi(4)=\frac{1}{48} (n-1) n^2 (n+1)^2 (n+2) (3 n^2+3 n-2)$$

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