find-k-1-n-k-C-n-k-2-interms-of-n-

Question Number 72389 by mathmax by abdo last updated on 28/Oct/19

f i n d \sum_{k = 1}^{n} k {(C_{n}^{k})}^{2} i n t e r m s o f n

Answered by mind is power last updated on 28/Oct/19

P (x) = {(1 + {te}^{ix})}^{n}

Q (x) = {(1 + {te}^{- ix})}^{n}

p (x) . Q (x) = {(t^{2} + 2 tcos (x) + 1)}^{n}

P (x) = \underset{k = 0}{\sum^{n}} C_{n}^{k} . t^{k} e^{ikx}

Q (x) = \underset{l = 0}{\sum^{n}} C_{n}^{l} t^{l} e^{- ilx}

P (x, t) . Q (x, t) = \underset{k = 0}{\sum^{n}} . \underset{l = 0}{\sum^{n}} . C_{n}^{k} . C_{n}^{l} t^{k} e^{i (k - l) x}

\int_{0}^{2 π} e^{i (k - l) x} dx = {\begin{cases} 0 if k \neq l \\ 2 π if k = l \end{cases}

\int_{0}^{2 π} P (x, t) . Q (x, t) dx = \int_{0}^{2 π} \underset{k = 0}{\sum^{n}} . \underset{l = 0}{\sum^{n}} . C_{n}^{k} . C_{n}^{l} t^{k + l} . e_{}^{i (k - l) x} dx

\int_{0}^{2 π} P (x, t) . Q (x, t) dx \sum_{k} \sum_{l} C_{n}^{k} . C_{n}^{l} \int_{0}^{2 π} t^{k + l} e^{i (k - l) x} dx = 2 π \underset{k = 1}{\sum^{n}} t^{2 k} {(C_{n}^{k})}^{2}

Q (t) = \int_{0}^{2 π} p (x, t) Q (x, t) dx = 2 π \underset{k = 1}{\sum^{n}} t^{2 k} {(C_{n}^{k})}^{2}

Q^{'} (t) ∣_{t = 1} = 2 π \underset{k = 1}{\sum^{n}} . {2 kt}^{2 k - 1} . {(C_{n}^{k})}^{2} ∣_{t = 1} = 4 π \underset{k = 1}{\sum^{n}} k {(C_{n}^{k})}^{2}

p (x, t) . Q (x, t) = {(t^{2} + 2 \cos (x) t + 1)}^{n}

Q^{'} (1) = \int_{0}^{2 π} . \frac{d}{dt} {(t^{2} + 2 \cos (x) t + 1)}^{n} ∣ t = 1 dx

= \int_{0}^{2 π} (2 t + 2 \cos (x)) n . {(t^{2} + 2 \cos (x) t + 1)}^{n - 1} ∣_{t = 1} dx

= n \int_{0}^{2 π} (2 + 2 \cos (x)) {(2 + 2 \cos (x))}^{n - 1} = {n 2}^{n} \int_{0}^{2 π} {(1 + \cos (x))}^{n} dx

1 + \cos (x) = {2 \cos}^{2} (\frac{x}{2}) - 1

= {n 2}^{n} \int_{0}^{2 π} (2^{n} \cos^{2 n} (\frac{x}{2})) dx \frac{x}{2} = t

= 2 n . 4^{n} \int_{0}^{π} \cos^{2 n} (t) dt = 4 n . 4^{n} \int_{0}^{\frac{π}{2}} \cos^{2 n} (t) dt =

by Walis integral we get

n . 4^{n + 1} \int_{0}^{\frac{π}{2}} \cos^{2 n} (t) dt = n . 4^{n + 1} W_{2 n} = 4 π . \underset{k = 1}{\sum^{n}} k {(C_{n}^{k})}^{2}

\Rightarrow \underset{k = 1}{\sum^{n}} k {(C_{n}^{k})}^{2} = \frac{n . 4^{n} W_{2 n}}{π}

Commented by mathmax by abdo last updated on 29/Oct/19

t h a n k y o u s i r .

Commented by mind is power last updated on 29/Oct/19

y^{'} re Welcom