Here is my approach of how to solve this problem.
∑ n = 1 N 2 − n z n = ∑ n = 1 N ( z 2 ) n = z 2 ( 1 − ( z 2 ) N ) 1 − z 2 \displaystyle \sum_{n=1}^{N}2^{-n}z^n = \sum_{n=1}^{N}\left(\frac{z}{2}\right)^n = \frac{\frac{z}{2}\left(1 - \left(\frac{z}{2}\right)^N\right)}{1 - \frac{z}{2}} n = 1 ∑ N 2 − n z n = n = 1 ∑ N ( 2 z ) n = 1 − 2 z 2 z ( 1 − ( 2 z ) N )
This result comes from the fact that the sum is a geometric series.
Now we have this sum:
∑ n = 1 10 2 − n sin ( n π 10 ) \displaystyle \sum_{n=1}^{10}2^{-n}\sin\left(\frac{n\pi}{10}\right) n = 1 ∑ 1 0 2 − n sin ( 1 0 n π )
which is the imaginary part of:
∑ n = 1 10 2 − n e i n π / 10 \displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10} n = 1 ∑ 1 0 2 − n e i n π / 1 0
Since
e i n π / 10 = cos ( n π 10 ) + i sin ( n π 10 ) \displaystyle e^{in\pi/10} = \cos\left(\frac{n\pi}{10}\right) + i\sin\left(\frac{n\pi}{10}\right) e i n π / 1 0 = cos ( 1 0 n π ) + i sin ( 1 0 n π ) ,
we can solve the sum that contains the exponential and then extract the imaginary part of it.
∑ n = 1 10 2 − n e i n π / 10 = ∑ n = 1 10 ( e i π / 10 2 ) n = e i π / 10 2 ( 1 − ( e i π / 10 2 ) 10 ) 1 − e i π / 10 2 \displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10} = \sum_{n=1}^{10}\left(\frac{e^{i\pi/10}}{2}\right)^n = \frac{\frac{e^{i\pi/10}}{2}\left(1 - \left(\frac{e^{i\pi/10}}{2}\right)^{10}\right)}{1 - \frac{e^{i\pi/10}}{2}} n = 1 ∑ 1 0 2 − n e i n π / 1 0 = n = 1 ∑ 1 0 ( 2 e i π / 1 0 ) n = 1 − 2 e i π / 1 0 2 e i π / 1 0 ( 1 − ( 2 e i π / 1 0 ) 1 0 ) = e i π / 10 2 ( 1 − ( e i π 2 10 ) ) 1 − e i π / 10 2 = e i π / 10 2 ( 2 10 2 10 − ( e i π 2 10 ) ) 2 2 − e i π / 10 2 = e i π / 10 ( 2 10 − e i π ) 2 10 ( 2 − e i π / 10 ) \displaystyle = \frac{\frac{e^{i\pi/10}}{2}\left(1 - \left(\frac{e^{i\pi}}{2^{10}}\right)\right)}{1 - \frac{e^{i\pi/10}}{2}} = \frac{\frac{e^{i\pi/10}}{2}\left(\frac{2^{10}}{2^{10}} - \left(\frac{e^{i\pi}}{2^{10}}\right)\right)}{\frac{2}{2} - \frac{e^{i\pi/10}}{2}}= \frac{e^{i\pi/10}\left(2^{10} - e^{i\pi}\right)}{2^{10}(2 - e^{i\pi/10})} = 1 − 2 e i π / 1 0 2 e i π / 1 0 ( 1 − ( 2 1 0 e i π ) ) = 2 2 − 2 e i π / 1 0 2 e i π / 1 0 ( 2 1 0 2 1 0 − ( 2 1 0 e i π ) ) = 2 1 0 ( 2 − e i π / 1 0 ) e i π / 1 0 ( 2 1 0 − e i π )
We know that
e i π = − 1 \displaystyle e^{i\pi} = -1 e i π = − 1 , then
∑ n = 1 10 2 − n e i n π / 10 = e i π / 10 ( 2 10 + 1 ) 2 10 ( 2 − e i π / 10 ) = ( 2 10 + 1 ) ( cos π / 10 + i sin π / 10 ) 2 10 ( 2 − cos π / 10 − i sin π / 10 ) \displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10} = \frac{e^{i\pi/10}\left(2^{10} + 1\right)}{2^{10}(2 - e^{i\pi/10})} = \frac{(2^{10} + 1)(\cos \pi/10 + i\sin \pi/10)}{2^{10}(2 - \cos \pi/10 - i\sin \pi/10)} n = 1 ∑ 1 0 2 − n e i n π / 1 0 = 2 1 0 ( 2 − e i π / 1 0 ) e i π / 1 0 ( 2 1 0 + 1 ) = 2 1 0 ( 2 − cos π / 1 0 − i sin π / 1 0 ) ( 2 1 0 + 1 ) ( cos π / 1 0 + i sin π / 1 0 )
Multiply the numerator and the denominator by
( 2 − cos π / 10 + i sin π / 10 ) \displaystyle (2 - \cos \pi/10 + i\sin \pi/10) ( 2 − cos π / 1 0 + i sin π / 1 0 ) n u m e r a t o r : \displaystyle \color{red} \bold{numerator:} n u m e r a t o r : ( cos π / 10 + i sin π / 10 ) ( 2 − cos π / 10 + i sin π / 10 ) = 2 cos ( π 10 ) − 1 + 2 i sin ( π 10 ) \displaystyle (\cos \pi/10 + i\sin \pi/10)(2 - \cos \pi/10 + i\sin \pi/10) = 2\cos\left(\frac{\pi}{10}\right) - 1 + 2i\sin\left(\frac{\pi}{10}\right) ( cos π / 1 0 + i sin π / 1 0 ) ( 2 − cos π / 1 0 + i sin π / 1 0 ) = 2 cos ( 1 0 π ) − 1 + 2 i sin ( 1 0 π ) d e n o m i n a t o r : \displaystyle \color{blue} \bold{denominator:} d e n o m i n a t o r : ( 2 − cos π / 10 − i sin π / 10 ) ( 2 − cos π / 10 + i sin π / 10 ) = 5 − 4 cos ( 10 π ) \displaystyle (2 - \cos \pi/10 - i\sin \pi/10)(2 - \cos \pi/10 + i\sin \pi/10) = 5 - 4\cos\left(\frac{10}{\pi}\right) ( 2 − cos π / 1 0 − i sin π / 1 0 ) ( 2 − cos π / 1 0 + i sin π / 1 0 ) = 5 − 4 cos ( π 1 0 )
This gives:
∑ n = 1 10 2 − n e i n π / 10 = ( 2 10 + 1 ) [ 2 cos ( π 10 ) − 1 + 2 i sin ( π 10 ) ] 2 10 [ 5 − 4 cos ( 10 π ) ] \displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10} = \frac{(2^{10} + 1)\bigg[2\cos\left(\frac{\pi}{10}\right) - 1 + 2i\sin\left(\frac{\pi}{10}\right)\bigg]}{2^{10}\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]} n = 1 ∑ 1 0 2 − n e i n π / 1 0 = 2 1 0 [ 5 − 4 cos ( π 1 0 ) ] ( 2 1 0 + 1 ) [ 2 cos ( 1 0 π ) − 1 + 2 i sin ( 1 0 π ) ]
Take the imaginary part of this result.
∑ n = 1 10 2 − n sin ( n π 10 ) = ( 2 10 + 1 ) [ 2 sin ( π 10 ) ] 2 10 [ 5 − 4 cos ( 10 π ) ] = 1025 [ 2 sin ( π 10 ) ] 1024 [ 5 − 4 cos ( 10 π ) ] = 1025 [ sin ( π 10 ) ] 512 [ 5 − 4 cos ( 10 π ) ] \displaystyle \sum_{n=1}^{10}2^{-n}\sin\left(\frac{n\pi}{10}\right) = \frac{(2^{10} + 1)\bigg[2\sin\left(\frac{\pi}{10}\right)\bigg]}{2^{10}\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]} = \frac{1025\bigg[2\sin\left(\frac{\pi}{10}\right)\bigg]}{1024\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]} = \frac{1025\bigg[\sin\left(\frac{\pi}{10}\right)\bigg]}{512\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]} n = 1 ∑ 1 0 2 − n sin ( 1 0 n π ) = 2 1 0 [ 5 − 4 cos ( π 1 0 ) ] ( 2 1 0 + 1 ) [ 2 sin ( 1 0 π ) ] = 1 0 2 4 [ 5 − 4 cos ( π 1 0 ) ] 1 0 2 5 [ 2 sin ( 1 0 π ) ] = 5 1 2 [ 5 − 4 cos ( π 1 0 ) ] 1 0 2 5 [ sin ( 1 0 π ) ] Finally. \displaystyle \textcolor{green}{\textbf{Finally.}} Finally. ∑ n = 1 10 2 − n sin ( n π 10 ) = 1025 sin ( π 10 ) 2560 − 2048 cos ( 10 π ) \displaystyle \sum_{n=1}^{10}2^{-n}\sin\left(\frac{n\pi}{10}\right) = \frac{1025\sin\left(\frac{\pi}{10}\right)}{2560 - 2048\cos\left(\frac{10}{\pi}\right)} n = 1 ∑ 1 0 2 − n sin ( 1 0 n π ) = 2 5 6 0 − 2 0 4 8 cos ( π 1 0 ) 1 0 2 5 sin ( 1 0 π )
Last edited: Jul 31, 2025
