show that ∀ n ∈N^∗ Σ_(k=1) ^n k(n−k)=(((n−1)(n+1))/6)


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Question Number 120480 by mathocean1 last updated on 31/Oct/20

$s h o w t h a t \forall n \in N^{*}$ $\sum_{k = 1}^{n} k (n - k) = \frac{(n - 1) (n + 1)}{6}$
	Answered by Ar Brandon last updated on 31/Oct/20

	$S_{k} = \underset{k = 1}{\sum^{n}} k (n - k) = \underset{k = 1}{\sum^{n}} (kn - k^{2})$ $= \frac{n^{2} (n + 1)}{2} - \frac{n (n + 1) (2 n + 1)}{6}$ $= \frac{{3 n}^{3} + {3 n}^{2} - ({2 n}^{3} + {3 n}^{2} + n)}{6}$ $= \frac{n^{3} - n}{6} = \frac{n (n - 1) (n + 1)}{6}$
	Answered by Dwaipayan Shikari last updated on 31/Oct/20

	$\underset{k = 1}{\sum^{n}} k n - \underset{k = 1}{\sum^{n}} k^{2}$ $= \frac{n^{2} (n + 1)}{2} - \frac{n (n + 1) (2 n + 1)}{6}$ $= \frac{n (n + 1)}{6} (3 n - 2 n - 1) = \frac{n (n^{2} - 1)}{6}$
	Commented by mathocean1 last updated on 31/Oct/20

	$t h a n k s s i r s$ $i s i t a l s o p o s s i b l e t o s h o w i t b y r e c u r r e n c e ?$
	Answered by Ar Brandon last updated on 31/Oct/20

	$Let P_{n} be the statement$ $\underset{k = 1}{\sum^{n}} k (n - k) = \frac{n (n - 1) (n + 1)}{6}$ $For n = 1, 0 = 0 (true)$ $For n = 2, (2 - 1) + 2 (2 - 2) = 1 = \frac{2 (1) (3)}{6} = 1 (true)$ $Suppose P_{n} true for all n \in N^{*} and show that P_{n + 1} is true .$ $\underset{k = 1}{\sum^{n}} k (n - k) = \frac{n (n - 1) (n + 1)}{6}$ $P_{n + 1} : \underset{k = 1}{\sum^{n + 1}} k (n - k) = \underset{k = 1}{\sum^{n}} k (n - k) - (n + 1)$ $= \frac{n (n - 1) (n + 1)}{6} - (n + 1) = \frac{(n + 1) (n^{2} - n - 6)}{6}$ $= \frac{(n + 1) (n - 3) (n + 2)}{6}$
	Commented by Ar Brandon last updated on 31/Oct/20
	I'm not getting it right.
	Answered by mathmax by abdo last updated on 31/Oct/20
	$Σ_(k=1) ^n k(n−k) =nΣ_(k=1) ^n k−Σ_(k=1) ^n k^2 =n((n(n+1))/2)−((n(n+1)(2n+1))/6) =((n(n+1))/2){n−((2n+1)/3)} =((n(n+1))/2)(((3n−2n−1)/3)) =((n(n+1)(n−1))/6)$
	$\sum_{k = 1}^{n} k (n - k) = n \sum_{k = 1}^{n} k - \sum_{k = 1}^{n} k^{2}$ $= n \frac{n (n + 1)}{2} - \frac{n (n + 1) (2 n + 1)}{6} = \frac{n (n + 1)}{2} {n - \frac{2 n + 1}{3}}$ $= \frac{n (n + 1)}{2} (\frac{3 n - 2 n - 1}{3}) = \frac{n (n + 1) (n - 1)}{6}$
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