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Question Number 82520 by peter frank last updated on 22/Feb/20

Commented by mathmax by abdo last updated on 22/Feb/20
$let S_n =Σ_(k=0) ^n 2^k cos(2kθ) ⇒S_n =Re(Σ_(k=0) ^n 2^k e^(i2kθ) ) =Re(Σ_(k=0) ^n (2 e^(i2θ) )^k ) but we have Σ_(k=0) ^n (2e^(2iθ) )^k =((1−(2e^(2iθ) )^(n+1) )/(1−2e^(2iθ) )) =((1−2^(n+1) e^(2(n+1)iθ) )/(1−2cos(2θ)−2isin(2θ))) =(((1−2cos(2θ)+2isin(2θ))(1−2^(n+1) e^(2(n+1)iθ) ))/((1−2cos(2θ))^2 +4sin^2 (2θ))) =(((1−2cos(2θ)+2isin(2θ)){1−2^(n+1) cos2(n+1)θ−2^(n+1) isin2(n+1)θ})/((1−2cos(2θ))^2 +4sin^2 (2θ))) =(((1−2cos(2θ))(1−2^(n+1) cos(2(n+1)θ)−i(1−2cos(2θ))2^(n+1) sin2(n+1)θ+2isin(2θ)(1−2^(n+1) cos2(n+1)θ)+2^(n+2) sin(2θ)sin2(n+1)θ)/((1−2cos(2θ))^2 +4sin^2 (2θ))) ⇒S_n =(((1−2cos(2θ))(1−2^(n+1) cos(2(n+1)θ)+2^(n+2) sin(2θ)sin2(n+1)θ)/(5−4cos(2θ)))$
$l e t S_{n} = \sum_{k = 0}^{n} 2^{k} c o s (2 k θ) \Rightarrow S_{n} = R e (\sum_{k = 0}^{n} 2^{k} e^{i 2 k θ})$ $= R e (\sum_{k = 0}^{n} {(2 e^{i 2 θ})}^{k}) b u t w e h a v e$ $\sum_{k = 0}^{n} {(2 e^{2 i θ})}^{k} = \frac{1 - {(2 e^{2 i θ})}^{n + 1}}{1 - 2 e^{2 i θ}} = \frac{1 - 2^{n + 1} e^{2 (n + 1) i θ}}{1 - 2 c o s (2 θ) - 2 i s i n (2 θ)}$ $= \frac{(1 - 2 c o s (2 θ) + 2 i s i n (2 θ)) (1 - 2^{n + 1} e^{2 (n + 1) i θ})}{{(1 - 2 c o s (2 θ))}^{2} + 4 {s i n}^{2} (2 θ)}$ $= \frac{(1 - 2 c o s (2 θ) + 2 i s i n (2 θ)) {1 - 2^{n + 1} c o s 2 (n + 1) θ - 2^{n + 1} i s i n 2 (n + 1) θ}}{{(1 - 2 c o s (2 θ))}^{2} + 4 {s i n}^{2} (2 θ)}$ $= \frac{(1 - 2 c o s (2 θ)) (1 - 2^{n + 1} c o s (2 (n + 1) θ) - i (1 - 2 c o s (2 θ)) 2^{n + 1} s i n 2 (n + 1) θ + 2 i s i n (2 θ) (1 - 2^{n + 1} c o s 2 (n + 1) θ) + 2^{n + 2} s i n (2 θ) s i n 2 (n + 1) θ}{{(1 - 2 c o s (2 θ))}^{2} + 4 {s i n}^{2} (2 θ)}$ $\Rightarrow S_{n} = \frac{(1 - 2 c o s (2 θ)) (1 - 2^{n + 1} c o s (2 (n + 1) θ) + 2^{n + 2} s i n (2 θ) s i n 2 (n + 1) θ}{5 - 4 c o s (2 θ)}$
	Answered by mr W last updated on 22/Feb/20
	$z=2(cos θ+i sin θ)=2e^(iθ) z^k =2^k (cos kθ+i sin kθ)=(2e^(iθ) )^k z^0 +z^1 +z^2 +...+z^n =Σ_(k=0) ^n (2e^(iθ) )^k ← sum of GP Σ_(k=0) ^n 2^k cos (kθ)+iΣ_(k=0) ^n 2^k sin (kθ)=(((2e^(iθ) )^(n+1) −1)/(2e^(iθ) −1)) Σ_(k=0) ^n 2^k cos (kθ)+iΣ_(k=0) ^n 2^k sin (kθ)=((2^(n+1) e^(i(n+1)θ) −1)/(2e^(iθ) −1)) Σ_(k=0) ^n 2^k cos (kθ)+iΣ_(k=0) ^n 2^k sin (kθ)=((2^(n+1) [cos (n+1)θ+i sin (n+1)θ]−1)/(2(cos θ+i sin θ)−1)) Σ_(k=0) ^n 2^k cos (kθ)+iΣ_(k=0) ^n 2^k sin (kθ)=(({[2^(n+1) cos (n+1)θ−1]+i [2^(n+1) sin (n+1)θ]}[(2cos θ−1)−i sin θ])/([(2cos θ−1)+i sin θ][(2cos θ−1)−i sin θ])) Σ_(k=0) ^n 2^k cos (kθ)+iΣ_(k=0) ^n 2^k sin (kθ)=(({[2^(n+1) cos (n+1)θ−1](2cos θ−1)+2^(n+1) sin (n+1)θ sin θ}+i {[2^(n+1) sin (n+1)θ](2cos θ−1)−[2^(n+1) cos (n+1)θ−1]sin θ})/((2cos θ−1)^2 +sin^2 θ)) ⇒Σ_(k=0) ^n 2^k cos (kθ)=(([2^(n+1) cos (n+1)θ−1](2cos θ−1)+2^(n+1) sin θ sin (n+1)θ)/((2cos θ−1)^2 +sin^2 θ)) ⇒Σ_(k=0) ^n 2^k sin (kθ)=((2^(n+1) (2 cos θ−1) sin (n+1)θ−[2^(n+1) cos (n+1)θ−1]sin θ)/((2cos θ−1)^2 +sin^2 θ))$
	$z = 2 (\cos θ + i \sin θ) = 2 e^{i θ}$ $z^{k} = 2^{k} (\cos k θ + i \sin k θ) = {(2 e^{i θ})}^{k}$ $z^{0} + z^{1} + z^{2} + . . . + z^{n} = \underset{k = 0}{\sum^{n}} {(2 e^{i θ})}^{k} \leftarrow s u m o f G P$ $\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{{(2 e^{i θ})}^{n + 1} - 1}{2 e^{i θ} - 1}$ $\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{2^{n + 1} e^{i (n + 1) θ} - 1}{2 e^{i θ} - 1}$ $\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{2^{n + 1} [\cos (n + 1) θ + i \sin (n + 1) θ] - 1}{2 (\cos θ + i \sin θ) - 1}$ $\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{{[2^{n + 1} \cos (n + 1) θ - 1] + i [2^{n + 1} \sin (n + 1) θ]} [(2 \cos θ - 1) - i \sin θ]}{[(2 \cos θ - 1) + i \sin θ] [(2 \cos θ - 1) - i \sin θ]}$ $\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{{[2^{n + 1} \cos (n + 1) θ - 1] (2 \cos θ - 1) + 2^{n + 1} \sin (n + 1) θ \sin θ} + i {[2^{n + 1} \sin (n + 1) θ] (2 \cos θ - 1) - [2^{n + 1} \cos (n + 1) θ - 1] \sin θ}}{{(2 \cos θ - 1)}^{2} + \sin^{2} θ}$ $\Rightarrow \underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) = \frac{[2^{n + 1} \cos (n + 1) θ - 1] (2 \cos θ - 1) + 2^{n + 1} \sin θ \sin (n + 1) θ}{{(2 \cos θ - 1)}^{2} + \sin^{2} θ}$ $\Rightarrow \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{2^{n + 1} (2 \cos θ - 1) \sin (n + 1) θ - [2^{n + 1} \cos (n + 1) θ - 1] \sin θ}{{(2 \cos θ - 1)}^{2} + \sin^{2} θ}$
	Commented by peter frank last updated on 22/Feb/20

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