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Question Number 107117 by Algoritm last updated on 08/Aug/20

	Answered by mr W last updated on 08/Aug/20

	$\frac{2 \sin 2 ° + 4 \sin 4 ° + . . . + 180 \sin 180 °}{9}$ $o r$ $\frac{2 \sin^{2} 2 ° + 4 \sin^{2} 4 ° + . . . + 180 \sin^{2} 180 °}{9} ?$
	Commented by Algoritm last updated on 08/Aug/20

	$2 \sin 2 °$
	Commented by mr W last updated on 08/Aug/20

	$\frac{2 \sin 2 ° + 4 \sin 4 ° + . . . + 180 \sin 180 °}{9} \approx 572.8996$ $\frac{2 \sin^{2} 2 ° + 4 \sin^{2} 4 ° + . . . + 180 \sin^{2} 180 °}{9} = 450$
	Answered by mathmax by abdo last updated on 08/Aug/20
	$let S =((2sin(2^o )+4sin(4^o )+....+180sin(180^o ))/9) ⇒ S =(1/9)Σ_(k=1) ^(90) 2ksin(2k^o ) but2 k^(o ) =2((kπ)/(180)) =((kπ)/(90))(radian) ⇒ S =(2/9)Σ_(k=0) ^(90) k sin(((kπ)/(90))) =(2/9) Im(Σ_(k=0) ^(90) k e^(i(((kπ)/(90)))) ) let S_n =Σ_(k=0) ^n k (e^((iπ)/(90)) )^k =f(e^((iπ)/(90)) ) with f(x) =Σ_(k=0) ^n k x^k we have Σ_(k=0) ^n x^k =((x^(n+1) −1)/(x−1)) (x≠1) ⇒Σ_(k=1) ^n kx^(k−1) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒Σ_(k=1) ^n kx^k =(x/((1−x)^2 )){ nx^(n+1) −(n+1)x^n +1} ⇒ Σ_(k=0) ^n k(e^((iπ)/(90)) )^k =(e^((iπ)/(90)) /((1−e^((iπ)/(90)) )^2 )){ n(e^((iπ)/(90)) )^(n+1) −(n+1)(e^((iπ)/(90)) )^n +1} ⇒ Σ_(k=0) ^(90) k(e^((iπ)/(90)) )^k =(e^((iπ)/(90)) /((1−cos((π/(90)))−isin((π/(90))))^2 )){ 90e^(i((91)/(90))π) +(91)+1} =((e^((iπ)/(90)) (1−cos((π/(90)))+isin((π/(90))))^2 )/((1−cos((π/(90)))^2 +sin^2 ((π/(90))))){ 90 cos(((91π)/(90)))+i90 sin(((91π)/(90)))+92} rest to extract im(of this quantity)....be continued...$
	$let S = \frac{2 \sin (2^{o}) + 4 \sin (4^{o}) + . . . . + 180 \sin (180^{o})}{9} \Rightarrow$ $S = \frac{1}{9} \sum_{k = 1}^{90} 2 ksin ({2 k}^{o}) but 2 k^{o} = 2 \frac{k π}{180} = \frac{k π}{90} (radian) \Rightarrow$ $S = \frac{2}{9} \sum_{k = 0}^{90} k \sin (\frac{k π}{90}) = \frac{2}{9} Im (\sum_{k = 0}^{90} k e^{i (\frac{k π}{90})})$ $let S_{n} = \sum_{k = 0}^{n} k {(e^{\frac{i π}{90}})}^{k} = f (e^{\frac{i π}{90}}) with f (x) = \sum_{k = 0}^{n} k x^{k} we have$ $\sum_{k = 0}^{n} x^{k} = \frac{x^{n + 1} - 1}{x - 1} (x \neq 1) \Rightarrow \sum_{k = 1}^{n} {kx}^{k - 1} = \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}}$ $\Rightarrow \sum_{k = 1}^{n} {kx}^{k} = \frac{x}{{(1 - x)}^{2}} {{nx}^{n + 1} - (n + 1) x^{n} + 1} \Rightarrow$ $\sum_{k = 0}^{n} k {(e^{\frac{i π}{90}})}^{k} = \frac{e^{\frac{i π}{90}}}{{(1 - e^{\frac{i π}{90}})}^{2}} {n {(e^{\frac{i π}{90}})}^{n + 1} - (n + 1) {(e^{\frac{i π}{90}})}^{n} + 1} \Rightarrow$ $\sum_{k = 0}^{90} k {(e^{\frac{i π}{90}})}^{k} = \frac{e^{\frac{i π}{90}}}{{(1 - \cos (\frac{π}{90}) - isin (\frac{π}{90}))}^{2}} {{90 e}^{i \frac{91}{90} π} + (91) + 1}$ $= \frac{e^{\frac{i π}{90}} {(1 - \cos (\frac{π}{90}) + isin (\frac{π}{90}))}^{2}}{(1 - \cos {(\frac{π}{90})}^{2} + \sin^{2} (\frac{π}{90})} {90 \cos (\frac{91 π}{90}) + i 90 \sin (\frac{91 π}{90}) + 92}$ $rest to extract im (of this quantity) . . . . be continued . . .$
	Commented by Algoritm last updated on 09/Aug/20

	$thanks$
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