prove that Σ_(n=1) ^r {n(n−(r/2))^2 }= r∙Σ_(n=1) ^(r/2) n^2 where r = 2k ; k ∈ N


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Question Number 24548 by Physics lover last updated on 20/Nov/17
$prove that Σ_(n=1) ^r {n(n−(r/2))^2 }= r∙Σ_(n=1) ^(r/2) n^2 where r = 2k ; k ∈ N$
$p r o v e t h a t$ $\underset{n = 1}{\sum^{r}} {n {(n - \frac{r}{2})}^{2}} = r \cdot \underset{n = 1}{\sum^{r / 2}} n^{2}$ $w h e r e r = 2 k; k \in N$
	Answered by jota last updated on 21/Nov/17

	$\underset{1}{\sum^{2 k}} n {(n - k)}^{2} =$ $Σ (n^{3} - 2 {k n}^{2} + k^{2}) = \frac{{(2 k)}^{2} {(2 k + 1)}^{2}}{4}$ $- 2 k \frac{2 k (2 k + 1) (4 k + 1)}{6} + k^{2} \frac{2 k (2 k + 1)}{2} =$ $= 2 k \frac{k (k + 1) (2 k + 1)}{6} = r \underset{1}{\sum^{r / 2 = k}} n^{2}$
	Commented by Physics lover last updated on 21/Nov/17

	$t h a n k y o u s o m u c h$
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