find Σ_(k=1) ^n k(C_n ^k )^2 interms of n


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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
	$P(x)=(1+te^(ix) )^n Q(x)=(1+te^(−ix) )^n p(x).Q(x)=(t^2 +2tcos(x)+1)^n P(x)=Σ_(k=0) ^n C_n ^k .t^k e^(ikx) Q(x)=Σ_(l=0) ^n C_n ^l t^l e^(−ilx) P(x,t).Q(x,t)=Σ_(k=0) ^n .Σ_(l=0) ^n .C_n ^k .C_n ^l t^k e^(i(k−l)x) ∫_0 ^(2π) e^(i(k−l)x) dx= { ((0 if k≠l)),((2π if k=l)) :} ∫_0 ^(2π) P(x,t).Q(x,t)dx=∫_0 ^(2π) Σ_(k=0) ^n .Σ_(l=0) ^n .C_n ^k .C_n ^l t^(k+l) .e_ ^(i(k−l)x) dx ∫_0 ^(2π) P(x,t).Q(x,t)dxΣ_k Σ_l C_n ^k .C_n ^l ∫_0 ^(2π) t^(k+l) e^(i(k−l)x) dx=2πΣ_(k=1) ^n t^(2k) (C_n ^k )^2 Q(t)=∫_0 ^(2π) p(x,t)Q(x,t)dx=2πΣ_(k=1) ^n t^(2k) (C_n ^k )^2 Q′(t)∣_(t=1) =2πΣ_(k=1) ^n .2kt^(2k−1) .(C_n ^k )^2 ∣_(t=1) =4πΣ_(k=1) ^n k(C_n ^k )^2 p(x,t).Q(x,t)=(t^2 +2cos(x)t+1)^n Q′(1)=∫_0 ^(2π) .(d/dt)(t^2 +2cos(x)t+1)^n ∣t=1 dx =∫_0 ^(2π) (2t+2cos(x))n.(t^2 +2cos(x)t+1)^(n−1) ∣_(t=1) dx =n∫_0 ^(2π) (2+2cos(x))(2+2cos(x))^(n−1) =n2^n ∫_0 ^(2π) (1+cos(x))^n dx 1+cos(x)=2cos^2 ((x/2))−1 =n2^n ∫_0 ^(2π) (2^n cos^(2n) ((x/2)))dx (x/2)=t =2n.4^n ∫_0 ^π cos^(2n) (t)dt=4n.4^n ∫_0 ^(π/2) cos^(2n) (t)dt= by Walis integral we get n.4^(n+1) ∫_0 ^(π/2) cos^(2n) (t)dt=n.4^(n+1) W_(2n) =4π.Σ_(k=1) ^n k(C_n ^k )^2 ⇒Σ_(k=1) ^n k(C_n ^k )^2 =((n.4^n W_(2n) )/π)$
	$P (x) = {(1 + {te}^{ix})}^{n}$ $Q (x) = {(1 + {te}^{- ix})}^{n}$ $p (x) . Q (x) = {(t^{2} + 2 tcos (x) + 1)}^{n}$ $P (x) = \underset{k = 0}{\sum^{n}} C_{n}^{k} . t^{k} e^{ikx}$ $Q (x) = \underset{l = 0}{\sum^{n}} C_{n}^{l} t^{l} e^{- ilx}$ $P (x, t) . Q (x, t) = \underset{k = 0}{\sum^{n}} . \underset{l = 0}{\sum^{n}} . C_{n}^{k} . C_{n}^{l} t^{k} e^{i (k - l) x}$ $\int_{0}^{2 π} e^{i (k - l) x} dx = {\begin{cases} 0 if k \neq l \\ 2 π if k = l \end{cases}$ $\int_{0}^{2 π} P (x, t) . Q (x, t) dx = \int_{0}^{2 π} \underset{k = 0}{\sum^{n}} . \underset{l = 0}{\sum^{n}} . C_{n}^{k} . C_{n}^{l} t^{k + l} . e_{}^{i (k - l) x} dx$ $\int_{0}^{2 π} P (x, t) . Q (x, t) dx \sum_{k} \sum_{l} C_{n}^{k} . C_{n}^{l} \int_{0}^{2 π} t^{k + l} e^{i (k - l) x} dx = 2 π \underset{k = 1}{\sum^{n}} t^{2 k} {(C_{n}^{k})}^{2}$ $Q (t) = \int_{0}^{2 π} p (x, t) Q (x, t) dx = 2 π \underset{k = 1}{\sum^{n}} t^{2 k} {(C_{n}^{k})}^{2}$ $Q^{'} (t) ∣_{t = 1} = 2 π \underset{k = 1}{\sum^{n}} . {2 kt}^{2 k - 1} . {(C_{n}^{k})}^{2} ∣_{t = 1} = 4 π \underset{k = 1}{\sum^{n}} k {(C_{n}^{k})}^{2}$ $p (x, t) . Q (x, t) = {(t^{2} + 2 \cos (x) t + 1)}^{n}$ $Q^{'} (1) = \int_{0}^{2 π} . \frac{d}{dt} {(t^{2} + 2 \cos (x) t + 1)}^{n} ∣ t = 1 dx$ $= \int_{0}^{2 π} (2 t + 2 \cos (x)) n . {(t^{2} + 2 \cos (x) t + 1)}^{n - 1} ∣_{t = 1} dx$ $= n \int_{0}^{2 π} (2 + 2 \cos (x)) {(2 + 2 \cos (x))}^{n - 1} = {n 2}^{n} \int_{0}^{2 π} {(1 + \cos (x))}^{n} dx$ $1 + \cos (x) = {2 \cos}^{2} (\frac{x}{2}) - 1$ $= {n 2}^{n} \int_{0}^{2 π} (2^{n} \cos^{2 n} (\frac{x}{2})) dx \frac{x}{2} = t$ $= 2 n . 4^{n} \int_{0}^{π} \cos^{2 n} (t) dt = 4 n . 4^{n} \int_{0}^{\frac{π}{2}} \cos^{2 n} (t) dt =$ $by Walis integral we get$ $n . 4^{n + 1} \int_{0}^{\frac{π}{2}} \cos^{2 n} (t) dt = n . 4^{n + 1} W_{2 n} = 4 π . \underset{k = 1}{\sum^{n}} k {(C_{n}^{k})}^{2}$ $\Rightarrow \underset{k = 1}{\sum^{n}} k {(C_{n}^{k})}^{2} = \frac{n . 4^{n} W_{2 n}}{π}$
	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

	$y^{'} re Welcom$
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