Question-36099

Question Number 36099 by rahul 19 last updated on 28/May/18

Commented by rahul 19 last updated on 01/Jun/18

Help in Q . 1 .

Commented by rahul 19 last updated on 31/Aug/18

Thank you prof Abdo :)

Commented by maxmathsup by imad last updated on 31/Aug/18

2) w e h a v e f (x + 1) = {(- 1)}^{x + 1} x - 2 f (x) x f r o m N f (1) = f (1986)

l e t f i n d S = f (1) + f (2) + \dots . f (1985)

w e h a v e \sum_{k = 1}^{1985} f (k + 1) = \sum_{k = 1}^{1985} k {(- 1)}^{k + 1} - 2 \sum_{k = 1}^{1985} f (k) \Rightarrow

\sum_{k = 2}^{1986} f (k) = \sum_{k = 1}^{1985} k {(- 1)}^{k + 1} - 2 S \Rightarrow \sum_{k = 1}^{1985} f (k) + f (1986) - f (1)

= - 2 S - \sum_{k = 1}^{1985} k {(- 1)}^{k} \Rightarrow 3 S = - \sum_{k = 1}^{1985} k {(- 1)}^{k} f o r x \neq 1

l e t p (x) = \sum_{k = 0}^{n} x^{k} \Rightarrow p^{^{'}} (x) = \sum_{k = 1}^{n} k x^{k - 1} \Rightarrow x p^{^{'}} (x) = \sum_{k = 1}^{n} k x^{k} 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

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

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

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

\sum_{k = 1}^{1985} k {(- 1)}^{k} = \frac{- (2 \times 1985 + 1) - 1}{4} = \frac{- 3972}{4} = - 993 \Rightarrow

3 S = 993 \Rightarrow S = 331 .

Commented by maxmathsup by imad last updated on 31/Aug/18

n e v e r m i n d s i r .