Question

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

• A. $24(1-i)$
• B. $48(1-i)$
• C. $24(1+i)$
• D. $48(1+i)$

Answer 1

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

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

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

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

$\therefore S=\sum _{p=1}^{32}(3p+2){\left[\sum _{q=1}^{10}(\sin \frac{29\pi }{11}-i\cos \frac{29\pi }{11})\right]}^p=\sum _{p=1}^{32}(3p+2)ip$

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

Now
$A=i+{2i}^2+{3i}^2+{3i}^3+...+{32i}^{32}\Rightarrow Ai=i^2+{2i}^3+...+{31i}^{32}+{32i}^{33}\Rightarrow A(1-i)=i+i^2+...+i^{32}-{32}^{33}$
$=\frac{i^{32}-1}{i-1}-32i=-32i[\because i^{32}=1]$

$\therefore A=\frac{-32i}{1-i}=\frac{32i(1+i)}{2}=16(1-i)$
and $B=i+i^2+...+i^{32}=0$
Hence $S=3\times 16(1-i)=48(1-i)$