Question

If $x_n=cos\left(\pi /2^n\right)+isin(\pi /2^n)$ then $x_1{x}_2{x}_3\dots$ infinity equal:

• A. $-i$
• B. $-1$
• C. $i$
• D. $1$

Answer 1

The correct option is B

$-1$

Explanation for the correct option.

We know that, $e^{i\theta }=\cos \theta +i\sin \theta$

So, $x_n=e^{i\mathrm{\pi }/2^{\mathrm{n}}}$

Now,

$\begin{array}{rcl}{x}_1{x}_2{x}_3\dots x_{\infty }& =& e^{i\mathrm{\pi }/2^1}\times e^{i\mathrm{\pi }/2^2}\times e^{i\mathrm{\pi }/2^3}\times .....\times e^{i\mathrm{\pi }/2^{\infty }}\\ & =& {e^{i\mathrm{\pi }/2}}^{\left(1+\frac{1}{2}+\frac{1}{2^n}+....+\frac{1}{2^{\infty }}\right)}\\ & =& {e^{i\mathrm{\pi }/2}}^{\left[\frac{1}{1-{\displaystyle \frac{1}{2}}}\right]}\left[\mathrm{sum}\mathrm{of}\mathrm{infinite}\mathrm{terms}\mathrm{of}\mathrm{G}.\mathrm{P}\right]\\ & =& e^{i\mathrm{\pi }}\\ & =& \mathrm{cos\pi }+i\mathrm{sin\pi }\\ & =& -1+0\\ & =& -1\end{array}$

Hence, option B is correct.