Question

$\sum _{p=1}^{32}(3p+2){\left[\sum _{q=1}^{10}(\sin \frac{2q\pi }{11}-i\cos \frac{2q\pi }{11})\right]}^p=$

• A. $8(1-i)$
• B. $16(1-i)$
• C. $48(1-i)$
• D. None of these

Answer 1

The correct option is C $48(1-i)$

$\sum _{q=1}^{10}\left(\sin \frac{2q\pi }{11}-i\sin \frac{2q\pi }{11}\right)$

$=\{\sin \frac{2\pi }{11}+\sin \frac{4\pi }{11}+...+10\}-i\{\cos \frac{2\pi }{11}+\cos \frac{4\pi }{11}+...+10\}$

$=\frac{\sin (\frac{2\pi }{11}+\frac{9\pi }{11})\sin \frac{10\pi }{11}}{\sin \frac{\pi }{11}}-i\frac{\cos (\frac{2\pi }{11}+\frac{9\pi }{11})\sin \frac{10\pi }{11}}{\sin \frac{\pi }{11}}$$=0-i(-1)=i$ ...(1)

$\therefore S=\sum _{p=1}^{32}(3q+2){\left[\sum _{q=1}^{10}\left(\sin \frac{2q\pi }{11}-i\cos \frac{2q\pi }{11}\right)\right]}^p$

$=\sum _{p=1}^{32}(3p+2)i^p=3\sum _{p=1}^{32}p{i}^p+2\sum _{p=1}^{32}{i}^p=3A+2B$ ...(2)

Now, $A=i+2i^2+3i^3+...+32i^{32}$$\Rightarrow Ai=i^2+2i^3+...+31i^{32}+32i^{33}$

$\Rightarrow A(1-i)=i+i^2+...+i^{32}-32i^{33}$$=\frac{i^{32}-1}{i-1}-32i=-32i[\because i^{32}=1]$

$\therefore A=\frac{-32i}{1-i}=\frac{32(1+i)}{2}=16(1-i)$ ...(3)

and, $B=i+i^2+...+i^{32}=0$ ...(4)

Hence, $S=3\times 16(1-i)=48(1-i).$