find-the-value-of-S-n-k-0-k-n-1-k-C-n-k-2k-1-

Question Number 27380 by abdo imad last updated on 05/Jan/18

f i n d t h e v a l u e o f S_{n} = \sum_{k = 0}^{k = n} \frac{{(- 1)}^{k} C_{n}^{k}}{2 k + 1} .

Commented by abdo imad last updated on 07/Jan/18

l e t i n t r o d u c e t h e p o l y n o m i a l p (x) = \sum_{k = 0}^{k = n} \frac{{(- 1)}^{k} C_{n}^{k}}{2 k + 1} x^{2 k + 1}

S_{n} = p (1) w e h a v e p^{,} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} x^{2 k}

= \sum_{k = 0}^{k = n} C_{n}^{k} {(- x^{2})}^{k} = {(1 - x^{2})}^{n}

p (x) = \int_{0}^{x} {(1 - t^{2})}^{n} d t + λ a n d λ = p (0) = 0

S_{n} = \int_{0}^{1} {(1 - t^{2})}^{n} d t a n d t h e c h . t = s i n x g i v e

S_{n} = \int_{0}^{\frac{π}{2}} {(1 - {s i n}^{2} x)}^{n} c o s x d x = \int_{0}^{\frac{π}{2}} {c o s x}^{2 n + 1} d x l e t p u t

I_{n} = \int_{0}^{\frac{π}{2}} {(c o s x)}^{n} d x (w a l l i s i n t e g r a l) i n t e g r a t i o n b y p a r t s

g i v e I_{n} = \frac{n - 1}{n} I_{n - 2} \Rightarrow I_{2 n + 1} = \frac{2 n}{2 n + 1} I_{2 n - 1}

\prod_{k = 1}^{n} I_{2 k + 1} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 1} \prod_{k = 1}^{n} I_{2 k - 1}

\Rightarrow I_{2 n + 1} = \frac{2^{n} (n!)}{3.5 \dots . (2 n + 1)} I_{1} b u t I_{1} = 1

S_{n} = \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!} .