Question

Value of ${\sum }_{k=0}^n{{}^{n}C}_k\sin \left(kx\right)\cos (n-k)x$ is

• A. $2^{n-1}\sin \left(nx\right)$
• B. $2^n\sin \left(nx\right)$
• C. $2^n\cos \left(nx\right)$
• D. none of these

Answer 1

Answer: (A)

$2^{n-1}\sin \left(nx\right)$

$S=\sum {}^n{C}_k\sin \left(kx\right)\cos (n-k)x$
Now writing the terms in reverse, we get
$S=\sum {}^n{C}_{n-k}\sin (n-k)x\cos \left(k\right)x$
Hence
$2S=\sum {}^n{C}_k\left[\sin \right(kx\left)\cos \right((n-k)x)+\cos (kx\left)\sin \right((n-k)x\left)\right]$
$=\sum {}^n{C}_k\sin \left(nx\right)$
Hence
$2S=2^n\sin \left(nx\right)$
$S=2^{n-1}\sin \left(nx\right)$