calcul en fonction de n Σ_(k=0) ^n 3^(k−1) (_k ^n ) Σ_(k=0) ^(k=n) sin(kx)(


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Question Number 163111 by SANOGO last updated on 03/Jan/22

$c a l c u l e n f o n c t i o n d e n$ $\underset{k = 0}{\sum^{n}} 3^{k - 1} (_{k}^{n})$ $\underset{k = 0}{\sum^{k = n}} s i n (k x) (_{k}^{n})$
	Answered by Mathspace last updated on 04/Jan/22
	$A_(n=Σ_(k=0) ^m C_n ^k sin(kx) ⇒) A_n =Im(Σ_(k=0) ^n C_n ^k e^(ikx) ) and Σ_(k=0) ^n C_n ^k (e^(ix) )^k =(1+e^(ix) )^n =(1+cosx +isinx)^n =(2cos^2 ((x/2))+2isin((x/2))cos((x/2)))^n =2^n cos^n ((x/2))(e^((ix)/2) )^n =2^n cos^n ((x/2)){cos(((nx)/2))+isin(((nx)/2))} ⇒ A_n =2^n cos^n ((x/2))sin(((nx)/2))$
	$A_{n = \sum_{k = 0}^{m} C_{n}^{k} s i n (k x) \Rightarrow}$ $A_{n} = I m (\sum_{k = 0}^{n} C_{n}^{k} e^{i k x}) a n d$ $\sum_{k = 0}^{n} C_{n}^{k} {(e^{i x})}^{k} = {(1 + e^{i x})}^{n}$ $= {(1 + c o s x + i s i n x)}^{n}$ $= {(2 {c o s}^{2} (\frac{x}{2}) + 2 i s i n (\frac{x}{2}) c o s (\frac{x}{2}))}^{n}$ $= 2^{n} {c o s}^{n} (\frac{x}{2}) {(e^{\frac{i x}{2}})}^{n}$ $= 2^{n} {c o s}^{n} (\frac{x}{2}) {c o s (\frac{n x}{2}) + i s i n (\frac{n x}{2})} \Rightarrow$ $A_{n} = 2^{n} {c o s}^{n} (\frac{x}{2}) s i n (\frac{n x}{2})$
	Commented by SANOGO last updated on 04/Jan/22

	$m e r c i b i e n$
	Answered by Mathspace last updated on 04/Jan/22

	$U_{n} = \sum_{k = 0}^{n} C_{n}^{k} 3^{k - 1} \Rightarrow$ $U_{n} = \frac{1}{3} \sum_{k = 0}^{n} C_{n}^{k} 3^{k}$ $= \frac{1}{3} {(3 + 1)}^{n} = \frac{4^{n}}{3}$
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