Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

$\rm \longrightarrow \large {(2)}^{ \frac{1}{4} } . {(4)}^{ \frac{1}{8} } . {(8)}^{ \frac{1}{16} } ... \infty$

$\rm \longrightarrow \large {(2)}^{ \frac{1}{4} } . {(2)}^{ \frac{2}{8} } . {(2)}^{ \frac{3}{16} } ... \infty$

$\rm \longrightarrow \large {(2)}^{ \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + ... \infty}$

$\rm \large {Let \: \: \: { \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + ... \infty} = s }$

$\rm \longrightarrow \large {(2)}^{ \large s}$

$\tt s = { \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} ... \infty} \\ \\ \tt \frac{s}{2} = \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + ... \infty \\ \\ \tt \rightarrow \bigg(s - \frac{s}{2} \bigg ) = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}...\infty\\ \\ \tt \rightarrow \frac{s}{2} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}...\infty \\ \\ \tt \rightarrow \frac{s}{2} = \dfrac{ \dfrac{1}{4} }{1 - \dfrac{1}{2} } \\ \\ \tt \rightarrow \frac{s}{2} = \dfrac{ \dfrac{1}{4} }{ \dfrac{1}{2} } \\ \\ \tt \rightarrow \frac{s}{2} = { \dfrac{1}{2} }\\ \\ \tt \rightarrow {s}= 1$

$\rm \longrightarrow \large {(2)}^{ \large s} = {2}^{1}$

Therefore, option (b) 2 is correct.

Answer 2

Answer:

option b is correct answer