Question

If $a=\underset{n\to \infty }{lim}\frac{\tan (\frac{\pi }{2^2}+\frac{\pi }{2^3}+....+\frac{\pi }{2^{n+1}})}{2^{n-1}}$, then value of $\pi a$ is

Answer 1

Given: $a={lim}_{n\to \infty }\frac{tan(\frac{\pi }{2^2}+\frac{\pi }{2^3}+.....\frac{\pi }{2^{n+1}})}{2^{n-1}}$

To find: Value of $\pi a$

Sol: $\frac{\pi }{2^2}(1+\frac{1}{2}+\frac{1}{2^2}+......\frac{1}{2^{n-1}})$is a geometric progression

$⟹\frac{\pi }{2^2}\left[\frac{1(1-(\frac{1}{2}{)}^n)}{1-\frac{1}{2}}\right]=(1-\frac{1}{2^n})\times \frac{\pi }{2}$

$\therefore {lim}_{n\to \infty }\frac{tan(\frac{\pi }{2}-\frac{\pi }{2^{n+1}})}{2^{n-1}}$

$a={lim}_{n\to \infty }\frac{\cot \left(\frac{\pi }{2^{n+1}}\right)}{2^{n-1}}$

$⟹a={lim}_{n\to \infty }\frac{1}{\tan \left(\frac{\pi }{2^{n+1}}\right)}.\frac{4}{2^{n-1}}$

$={lim}_{n\to \infty }\frac{\left(\frac{\pi }{2^{n+1}}\right)}{\tan \left(\frac{\pi }{2^{n+1}}\right)}.\frac{1}{\left(\frac{\pi }{2^{n+1}}\right).2^{n+1}}=\frac{4}{\pi }$

$⟹a\pi =4$

Hence, correct answer is $4$