STD 11 - 8. sequence and series — JEE STD 11 Science — Question

$ = {2^{1/4 + 2/8 + 3/16 + ......}} = {2^S}$, where $S$ is given by

$S = \frac{1}{4} + 2\frac{1}{8} + 3\frac{1}{{16}} + 4\frac{1}{{32}} + ...........\infty $......$(i)$

$ \Rightarrow $$\frac{1}{2}S = {\rm{ }}\frac{1}{8} + \frac{2}{{16}} + \frac{3}{{32}} + \frac{4}{{64}} + ..........\infty $......$(ii)$

Subtracting $(ii)$ from $(i),$ we get $S = 1$.

Hence required product $ = {2^1} = 2$.