cslculate-k-1-n-k-2-n-1-k-

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}

\dots \dots \dots \dots \dots \dots \dots . .

\dots \dots \dots \dots \dots \dots \dots \dots

T_{n} = (n + 1) n^{2} - n^{3}

s o T_{1} + T_{2} + \dots + 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 \dots