Question

$\sum _{n=1}^{\infty }\frac{1}{\left(2^{n+1}\right)\left(2^{n+3}\right)}$

Answer 1

According to question,

$\frac{1}{\left(2^{n+1}\right)\left(2^{n+3}\right)}$

$⟹\frac{1}{2^{2(n+2)}}$

$⟹\frac{1}{4^{n+2}}$

On putting limits , we get series as,

$⟹\frac{1}{4^3}+\frac{1}{4^4}+\frac{1}{4^5}\dots \dots \dots ..$

Which is a sum of an infinite GP,

Hence, Summation= $\frac{\frac{1}{4^3}}{1-\frac{1}{4}}$

$⟹\frac{1}{48}$