cslculate Σ_(k=1) ^n k^2 (n+1−k)


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Question Number 35036 by abdo mathsup 649 cc last updated on 14/May/18

$c s l c u l a t e \sum_{k = 1}^{n} k^{2} (n + 1 - k)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18
	$T_k =(n+1)k^2 −k^3 T_1 =(n+1)1^2 −1^3 T_2 =(n+1)2^2 −2^3 T_3 =(n+1)3^2 −3^3 ....................... ........................ T_n =(n+1)n^2 −n^3 so T_1 +T_2 +...+T_n =(n+1)((n(n+1)(2n+1))/6)−{((n(n+1))/2)}^2$
	$T_{k} = (n + 1) k^{2} - k^{3}$ $T_{1} = (n + 1) 1^{2} - 1^{3}$ $T_{2} = (n + 1) 2^{2} - 2^{3}$ $T_{3} = (n + 1) 3^{2} - 3^{3}$ $. . . . . . . . . . . . . . . . . . . . . . .$ $. . . . . . . . . . . . . . . . . . . . . . . .$ $T_{n} = (n + 1) n^{2} - n^{3}$ $s o T_{1} + T_{2} + . . . + T_{n}$ $= (n + 1) \frac{n (n + 1) (2 n + 1)}{6} - {\frac{n (n + 1)}{2}}^{2}$
	Commented by math khazana by abdo last updated on 15/May/18

	$c o r r e c t a n s w e r t h a n k s . . .$
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