find S_n =Σ_(k=0) ^n k^2 C_n ^k cos(2kx) interms of n.


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Question Number 56962 by turbo msup by abdo last updated on 27/Mar/19

$f i n d S_{n} = \sum_{k = 0}^{n} k^{2} C_{n}^{k} c o s (2 k x)$ $i n t e r m s o f n .$
Commented by maxmathsup by imad last updated on 29/Mar/19
$we have S_n =Re( Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) )=Re(W_n ) let p(x)=Σ_(k=0) ^n C_n ^k e^(i2kx) =Σ_(k=0) ^n C_n ^k (e^(i2x) )^k =(1+e^(i2x) )^n but p^′ (x)=2i Σ_(k=0) ^n k C_n ^k e^(i2kx) and p^(′′) (x)=−4 Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) ⇒ Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) =−(1/4)p^((2)) (x) we have p(x)=(e^(i2x) +1)^n ⇒ p^′ (x)=2in(e^(i2x) +1)^(n−1) and p^((2)) (x)=−4n(n−1)(e^(i2x) +1)^(n−2) but (e^(i2x) +1)^(n−2) =(cos(2x)+isin(2x)+1)^(n−2) =(2cos^2 (x)+2isinx cosx)^(n−2) =(2cosx)^(n−2) (e^(ix) )^(n−2) =(2cosx)^(n−2) e^(i(n−2)x) =(2cosx)^(n−2) {cos(n−2)x +isin(n−2)x} ⇒ Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) =−(1/4)(−4n(n−1)) (2cosx)^(n−2) {cos(n−2)x +isin(n−2)x} =n(n−1)2^(n−2) cos^(n−2) x{cos(n−2)x +isin(n−2)x} ⇒ S_n =n(n−1)2^(n−2) cos^(n−2) (x) cos((n−2)x) .$
$w e h a v e S_{n} = R e (\sum_{k = 0}^{n} k^{2} C_{n}^{k} e^{i 2 k x}) = R e (W_{n})$ $l e t p (x) = \sum_{k = 0}^{n} C_{n}^{k} e^{i 2 k x} = \sum_{k = 0}^{n} C_{n}^{k} {(e^{i 2 x})}^{k} = {(1 + e^{i 2 x})}^{n}$ $b u t p^{^{'}} (x) = 2 i \sum_{k = 0}^{n} k C_{n}^{k} e^{i 2 k x} a n d p^{^{″}} (x) = - 4 \sum_{k = 0}^{n} k^{2} C_{n}^{k} e^{i 2 k x} \Rightarrow$ $\sum_{k = 0}^{n} k^{2} C_{n}^{k} e^{i 2 k x} = - \frac{1}{4} p^{(2)} (x) w e h a v e p (x) = {(e^{i 2 x} + 1)}^{n} \Rightarrow$ $p^{^{'}} (x) = 2 i n {(e^{i 2 x} + 1)}^{n - 1} a n d p^{(2)} (x) = - 4 n (n - 1) {(e^{i 2 x} + 1)}^{n - 2}$ $b u t {(e^{i 2 x} + 1)}^{n - 2} = {(c o s (2 x) + i s i n (2 x) + 1)}^{n - 2}$ $= {(2 {c o s}^{2} (x) + 2 i s i n x c o s x)}^{n - 2} = {(2 c o s x)}^{n - 2} {(e^{i x})}^{n - 2} = {(2 c o s x)}^{n - 2} e^{i (n - 2) x}$ $= {(2 c o s x)}^{n - 2} {c o s (n - 2) x + i s i n (n - 2) x} \Rightarrow$ $\sum_{k = 0}^{n} k^{2} C_{n}^{k} e^{i 2 k x} = - \frac{1}{4} (- 4 n (n - 1)) {(2 c o s x)}^{n - 2} {c o s (n - 2) x + i s i n (n - 2) x}$ $= n (n - 1) 2^{n - 2} {c o s}^{n - 2} x {c o s (n - 2) x + i s i n (n - 2) x} \Rightarrow$ $S_{n} = n (n - 1) 2^{n - 2} {c o s}^{n - 2} (x) c o s ((n - 2) x) .$
	Answered by Smail last updated on 27/Mar/19

	$S_{n} (x) = \underset{k = 0}{\sum^{n}} k^{2} C_{n}^{k} c o s (2 k x) = R e (\underset{k = 0}{\sum^{n}} k^{2}^{n} C_{k} e^{2 i k x})$ $z^{″} (x) = \underset{k = 0}{\sum^{n}} k^{2}^{n} C_{k} e^{2 i k x}$ $z^{'} (x) = \frac{1}{2 i} \underset{k = 0}{\sum^{n}} k^{n} C_{k} e^{2 i k x} + c$ $z (x) = - \frac{1}{4} \underset{k = 0}{\sum^{n}}^{n} C_{k} e^{2 i k x} + c x + a$ $= \frac{- 1}{4} {(1 + e^{2 i x})}^{n} + c x + a$ $z^{'} (x) = \frac{1}{2 i} \times {n e}^{2 i x} {(1 + e^{2 i x})}^{n - 1} + c$ $z^{″} (x) = n (e^{2 i x} {(1 + e^{2 i x})}^{n - 1} + (n - 1) e^{4 i x} {(1 + e^{2 i x})}^{n - 2})$ $= {n e}^{2 i x} {(1 + e^{2 i x})}^{n - 2} (1 + e^{2 i x} + (n - 1) e^{2 i x})$ $= {n e}^{2 i x} {(1 + e^{2 i x})}^{n - 2} (1 + {n e}^{2 i x})$ $= {n e}^{2 i x} {(1 + c o s (2 x) + i s i n (2 x))}^{n - 2} (1 + {n e}^{2 i x})$ $= {n e}^{2 i x} {(2 {c o s}^{2} x + 2 i s i n x c o s x)}^{n - 2} (1 + {n e}^{2 i x})$ $= {n e}^{2 i x} 2^{n - 2} {c o s}^{n - 2} (x) {(e^{i x})}^{n - 2} (1 + {n e}^{2 i x})$ $= 2^{n - 2} {n c o s}^{n - 2} (x) e^{i n x} (1 + {n e}^{2 i x})$ $= 2^{n - 2} {n c o s}^{n - 2} x (e^{i n x} + {n e}^{i (n + 2) x})$ $= 2^{n - 2} {n c o s}^{n - 2} x (c o s n x + i s i n (n x) + n c o s ((n + 2) x) + i s i n ((n + 2) x))$ $S_{n} = 2^{n - 2} {n c o s}^{n - 2} x (c o s n x + n c o s (n + 2) x)$
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