calculate-B-n-k-0-n-1-k-2k-2-1-interms-of-n-

Question Number 37891 by abdo mathsup 649 cc last updated on 19/Jun/18

c a l c u l a t e B_{n} = \sum_{k = 0}^{n} {(- 1)}^{k} (2 k^{2} + 1) i n t e r m s o f n .

Commented by prof Abdo imad last updated on 24/Jun/18

B_{n} = 2 \sum_{k = 0}^{n} k^{2} {(- 1)}^{k} + \sum_{k = 0}^{n} {(- 1)}^{k}

l e t p (x) = \sum_{k = 0}^{n} x^{k} w i t h x \neq 1

p^{^{'}} (x) = \sum_{k = 1}^{n} k x^{k - 1} \Rightarrow x p^{^{'}} (x) = \sum_{k = 1}^{n} {k x}^{k} \Rightarrow

\sum_{k = 1}^{n} k^{2} x^{k - 1} = p^{^{'}} (x) + {x p}^{″} (x) \Rightarrow

\sum_{k = 1}^{n} k^{2} x^{k} = {x p}^{^{'}} (x) + x^{2} p^{″} (x) b u t

p (x) = \frac{x^{n + 1} - 1}{x - 1} \Rightarrow p^{^{'}} (x) = \frac{n x^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}}

p^{^{″}} (x) = \frac{(n (n + 1) x^{n} - n (n + 1) x^{n - 1}) {(x - 1)}^{2} - 2 (x - 1) ({n x}^{n + 1} - (n + 1) x^{n} + 1)}{{(x - 1)}^{4}}

= \frac{(x - 1) n (n + 1) (x^{n} - x^{n - 1}) - 2 ({n x}^{n + 1} - (n + 1) x^{n} + 1)}{{(x - 1)}^{3}}

\sum_{k = 1}^{n} k^{2} {(- 1)}^{k} = - p^{^{'}} (- 1) + p^{^{″}} (- 1)

= - \frac{n {(- 1)}^{n + 1} - (n + 1) {(- 1)}^{n} + 1}{4} + \frac{- 2 n (n + 1) ({(- 1)}^{n} - {(- 1)}^{n - 1}) - 2 {n {(- 1)}^{n + 1} - (n + 1) {(- 1)}^{n} + 1}}{- 8}

a l s o \sum_{k = 0}^{n} {(- 1)}^{k} = \frac{1 - {(- 1)}^{n + 1}}{2}

s o t h e v a l u e o f S i s d e t e r m i n e d .