calculate-A-n-k-0-n-1-k-2k-3-interms-of-n-

Question Number 37890 by abdo mathsup 649 cc last updated on 19/Jun/18

c a l c u l a t e A_{n} = \sum_{k = 0}^{n} {(- 1)}^{k} (2 k + 3) i n t e r m s o f n

Commented by prof Abdo imad last updated on 20/Jun/18

A_{n} = 2 \sum_{k = 0}^{n} k {(- 1)}^{k} + 3 \sum_{k = 0}^{n} {(- 1)}^{k} b u t

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

l e t p (x) = \sum_{k = 0}^{n} x^{k} w e h a v e p (x) = \frac{x^{n + 1} - 1}{x - 1}

i f x \neq - 1 a n d 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 {(- 1)}^{k} = - p^{^{'}} (- 1)

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

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

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

p^{^{'}} (- 1) = \frac{n {(- 1)}^{n + 1} - (n + 1) {(- 1)}^{n} + 1}{4}

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

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

A_{n} = 2 \frac{(2 n + 1) {(- 1)}^{n} - 1}{4} + \frac{3}{2} (1 + {(- 1)}^{n})

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

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

A_{n} = (n + 2) {(- 1)}^{n} + 1 .

Commented by math khazana by abdo last updated on 20/Jun/18

x \neq 1