let-u-n-k-0-n-3k-1-1-k-1-calculate-interms-of-n-S-n-u-0-u-1-u-2-u-n-2-calculate-u-0-u-1-u-2-u-57-

Question Number 40370 by math khazana by abdo last updated on 20/Jul/18

l e t u_{n} = \sum_{k = 0}^{n} (3 k + 1) {(- 1)}^{k}

1) c a l c u l a t e i n t e r m s o f n

S_{n} = u_{0} + u_{1} + u_{2} + \dots . + u_{n}

2) c a l c u l a t e u_{0} + u_{1} + u_{2} + \dots . + u_{57}

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

1) w e h a v e u_{n} = 3 \sum_{k = 0}^{n} k {(- 1)}^{k} + \sum_{k = 0}^{n} {(- 1)}^{k} b u t

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

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

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) 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}} \Rightarrow

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

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

u_{n} = \frac{3}{4} {(2 n + 1) {(- 1)}^{n} - 1} + \frac{1 + {(- 1)}^{n}}{2} w e h a v e

S_{n} = \sum_{k = 0}^{n} u_{k}

= \frac{3}{4} \sum_{k = 0}^{n} (2 k + 1) {(- 1)}^{k} - \frac{3}{4} \sum_{k = 0}^{n} (1)

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

= \frac{3}{2} \sum_{k = 0}^{n} k {(- 1)}^{k} + \frac{3}{4} \sum_{k = 0}^{n} {(- 1)}^{k} - \frac{3}{4} (n + 1)

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

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

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

S_{n} = \frac{3}{8} {1 - (2 n + 1) {(- 1)}^{n}} + \frac{5}{8} {1 + {(- 1)}^{n}}

- \frac{n + 1}{4} .