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Question Number 143461 by Willson last updated on 14/Jun/21

Prove that : \forall n \in N^{*}

a . \underset{k = 1}{\sum^{n}} C_{n}^{k} \frac{{(- 1)}^{k}}{k} = \underset{k = 1}{\sum^{n}} \frac{1}{k}

b . \underset{k = 1}{\sum^{n}} C_{n}^{k} \frac{{(- 1)}^{k}}{2 k + 1} = \frac{4^{n}}{(2 n + 1) C_{2 n}^{n}}

Answered by Dwaipayan Shikari last updated on 14/Jun/21

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

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

= \int_{0}^{1} \frac{{(1 - u)}^{n} - 1}{u} d u = - \int_{0}^{1} \frac{1 - u^{n}}{1 - u} d u = - H_{n}

Answered by mathmax by abdo last updated on 14/Jun/21

1) let p (x) = \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k}}{k} x^{k} \Rightarrow p^{^{'}} (x) = \sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{k} x^{k - 1}

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

= \frac{1}{x} ({(1 - x)}^{n} - 1) \Rightarrow p (x) = \int_{0}^{x} \frac{{(1 - t)}^{n} - 1}{t} dt + K

K = p (0) = 0 \Rightarrow p (x) = \int_{0}^{x} \frac{(1 - t - 1) (1 + (1 - t) + {(1 - t)}^{2} + \dots . + {(1 - t)}^{n - 1})}{t} dt

= - \int_{0}^{x} \sum_{k = 0}^{n - 1} {(1 - t)}^{k} dt = \sum_{k = 0}^{n - 1} {[\frac{1}{k + 1} {(1 - t)}^{k + 1}]}_{0}^{x}

= \sum_{k = 0}^{n - 1} \frac{1}{k + 1} ({(1 - x)}^{k + 1} - 1) = p (x) \Rightarrow

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

Answered by mathmax by abdo last updated on 15/Jun/21

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

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

p (x) = \int_{0}^{x} ({(1 - t^{2})}^{n} - 1) dt + K

k = p (0) = 0 \Rightarrow p (x) = \int_{0}^{x} {(1 - t^{2})}^{n} dt - x

\Rightarrow p (1) = \int_{0}^{1} {(1 - t^{2})}^{n} dt - 1

\int_{0}^{1} {(1 - t^{2})}^{n} dt =_{t = sinx} \int_{0}^{\frac{π}{2}} \cos^{2 n} x cosx dx

= \int_{0}^{\frac{π}{2}} \cos^{2 n + 1} xdx = \int_{0}^{\frac{π}{2}} \cos^{2 n + 1} x \sin^{o} x dx

2 p - 1 = 2 n + 1 \Rightarrow 2 p = 2 n + 2 \Rightarrow p = n + 1

2 q - 1 = 0 \Rightarrow q = \frac{1}{2} \Rightarrow

\int_{0}^{\frac{π}{2}} \cos^{2 n + 1} xdx = \int_{0}^{\frac{π}{2}} \cos^{2 (n + 1) - 1} \sin^{2 \frac{1}{2} - 1} xdx

= \frac{1}{2} B (n + 1, \frac{1}{2}) = \frac{1}{2} \frac{Γ (n + 1) Γ (\frac{1}{2})}{Γ (n + \frac{3}{2})} = \frac{n! \sqrt{π}}{2 Γ (n + \frac{3}{2})}

. . be continued \dots .