Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

$\\ \bigstar \large \underline{ \underline{\bold{Solution :- }}}$

$\\ \sf : \: 1 + \frac{1}{4(2!)} + \frac{1}{16(4 ! )} + \frac{1}{64(6 ! )} + ... \infty$

$\\ \sf \implies\: 1 + \frac{1}{ {2}^{2} (2!)} + \frac{1}{ {2}^{4} (4 ! )} + \frac{1}{ {2}^{6} (6 ! )} + ... \infty$

$\\ \sf \implies\: \frac{1}{2} \bigg[2 \bigg(1 + \frac{1}{ {2}^{2} (2!)} + \frac{1}{ {2}^{4} (4 ! )} + \frac{1}{ {2}^{6} (6 ! )} + ... \infty \bigg)\bigg ]$

$\\ \sf \implies\: \frac{1}{2} \bigg[2 \bigg(1 + \frac{ {x}^{2} }{ 2!} + \frac{ {x}^{4} }{ 4 ! } + \frac{ {x}^{6} }{ 6 ! } + ... \infty \bigg)\bigg ]$

$\\ \sf \implies\: \frac{1}{2} [ {e}^{x} + {e}^{ - x} ] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (when \: x = \frac{1}{2} )$

$\\ \sf \implies\: \frac{1}{2} [ {e}^{ \frac{1}{2} } + {e}^{ - \frac{1}{2} } ] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\\ \sf \implies \: \frac{e + 1}{2 \sqrt{e} }$

Answer 2

Answer:

option a is correct answer is