calculate-k-0-n-1-k-2k-1-C-n-k-

Question Number 97984 by abdomathmax last updated on 10/Jun/20

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

Answered by mr W last updated on 11/Jun/20

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

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

\int_{0}^{1} {(1 - x^{2})}^{n} d x = \underset{k = 0}{\sum^{n}} C_{k}^{n} {(- 1)}^{k} \int_{0}^{1} x^{2 k} d x

\int_{0}^{1} {(1 - x^{2})}^{n} d x = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\int_{0}^{\frac{π}{2}} {(1 - \sin^{2} t)}^{n} \cos t d t = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\int_{0}^{\frac{π}{2}} \cos^{2 n + 1} t d t = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\frac{2 n}{2 n + 1} \times \frac{2 n - 2}{2 n - 1} \times \dots \times \frac{2}{3} \times \int_{0}^{\frac{π}{2}} \cos t d t = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\frac{2 n}{2 n + 1} \times \frac{2 n - 2}{2 n - 1} \times \dots \times \frac{2}{3} \times 1 = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\frac{(2 n)!!}{(2 n + 1)!!} = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\frac{{[(2 n)!!]}^{2}}{(2 n + 1)!} = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\frac{{[2^{n} n!]}^{2}}{(2 n + 1)!} = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n}

\Rightarrow \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} C_{k}^{n} = \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!}

Commented by mathmax by abdo last updated on 10/Jun/20

thank you sir mrw

Answered by mathmax by abdo last updated on 11/Jun/20

let p (x) = \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 1} C_{n}^{k} x^{2 k + 1} we have

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

= {(- 1)}^{n} {(x^{2} - 1)}^{n} = {(1 - x^{2})}^{n} \Rightarrow p (x) = \int_{0}^{x} {(1 - t^{2})}^{n} dt + c (c = 0) and S_{n} = p (1)

= \int_{0}^{1} {(1 - t^{2})}^{n} dt =_{t = \sin θ} \int_{0}^{\frac{π}{2}} {(\cos^{2} θ)}^{n} \cos θ d θ = \int_{0}^{\frac{π}{2}} \cos^{2 n + 1} θ d θ =

let U_{n} = \int_{0}^{\frac{π}{2}} \cos^{n} t dt \Rightarrow U_{n} = \int_{0}^{\frac{π}{2}} \cos^{n - 2} t (1 - \sin^{2} t) dt

= U_{n - 2} - \int_{0}^{\frac{π}{2}} \sin^{2} t \cos^{n - 2} t dt but by parts f^{^{'}} = sint \cos^{n - 2} and g = sint

\int_{0}^{\frac{π}{2}} sint ({sintcos}^{n - 2} t) dt = {[- \frac{1}{n - 1} \cos^{n - 1} t sint]}_{0}^{\frac{π}{2}} + \frac{1}{n - 1} \int_{0}^{\frac{π}{2}} cost \cos^{n - 1} t dt

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

\frac{n}{n - 1} U_{n} = U_{n - 2} \Rightarrow U_{n} = \frac{n - 1}{n} U_{n - 2} \Rightarrow U_{2 n + 1} = \frac{2 n}{2 n + 1} U_{2 n - 1} \Rightarrow

\prod_{k = 1}^{n} U_{2 k + 1} = \prod_{k = 1}^{n} \frac{2 k}{(2 k + 1)} \prod_{k = 1}^{n} U_{2 k - 1} \Rightarrow

U_{2 n + 1} = U_{1} \times \prod_{k = 1}^{n} \frac{2 k}{(2 k + 1)} (U_{1} = 1) \Rightarrow S_{n} = \int_{0}^{\frac{π}{2}} \cos^{2 n + 1} t dt = \prod_{k = 1}^{n} \frac{2 k}{(2 k + 1)}