Question

Any complex number in the polar form can be expressed in Euler's form as $\cos \theta +i\sin \theta =e^{i\theta }$. This form of the complex number is useful in finding the sum of series $\sum _{r=0}^n{}^n{C}_r(\cos \theta +i\sin \theta {)}^r$.
$\sum _{r=0}^n{}^n{C}_r(\cos r\theta +i\sin r\theta )=\sum _{r=0}^n{}^n{C}_r{e}^{ir\theta }=\sum _{r=0}^n{}^n{C}_r(e^{i\theta }{)}^r=(1+e^{i\theta }{)}^n$
Also, we know that the sum of binomial series does not change if $r$ is replaced by $n-r$. Using these facts, answer the following questions.

The value of $\sum _{r=0}^{100}{}^{100}{C}_r\left(\sin rx\right)$ is equal to

• A. $2^{100}{\cos }^{100}\left(\frac{x}{2}\right)\sin 50x$
• B. $2^{100}\sin \left(50x\right)\cos \left(\frac{x}{2}\right)$
• C. $2^{101}{\cos }^{100}\left(50x\right)\sin \left(\frac{x}{2}\right)$
• D. $2^{101}{\sin }^{100}\left(50x\right)\cos \left(50x\right)$

Answer 1

The correct option is A $2^{100}{\cos }^{100}\left(\frac{x}{2}\right)\sin 50x$

$\sum _{r=0}^{100}{}^{100}{C}_r\left(\sin rx\right)=\text{Im}\left(\sum _{r=0}^{100}{}^{100}{C}_r{e}^{irx}\right)$
$=\text{Im}\left(\sum _{r=0}^{100}{}^{100}{C}_r\right(e^{ix}{)}^r)$
$=\text{Im}{(1+e^{ix})}^{100}$
$=\text{Im}(1+\cos x+i\sin x{)}^{100}$
$=\text{Im}{(2{\cos }^2\frac{x}{2}+2i\sin \frac{x}{2}\cos \frac{x}{2})}^{100}$
$=\text{Im}{\left(2\cos \frac{x}{2}(\cos \frac{x}{2}+i\sin \frac{x}{2})\right)}^{100}$
$=2^{100}{\cos }^{100}\left(\frac{x}{2}\right)\sin 50x$