Question

Elaborate all the steps involved

Answer 1

Given Question :- The sum of the series is

$\rm \: 1 + \dfrac{1}{4.2!} + \dfrac{1}{16.4!} + \dfrac{1}{64.6!} + \cdots \cdots \: \infty$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: 1 + \dfrac{1}{4.2!} + \dfrac{1}{16.4!} + \dfrac{1}{64.6!} + \cdots \cdots \: \infty$

We know, from exponential series

$\red{\rm \: {e}^{x} = 1 + x + \frac{ {x}^{2} }{2!} + \frac{ {x}^{3} }{3!} + \frac{ {x}^{4} }{4!} + \cdots \cdots}$

and

$\red{\rm \: {e}^{ - x} = 1 - x + \frac{ {x}^{2} }{2!} - \frac{ {x}^{3} }{3!} + \frac{ {x}^{4} }{4!} +\cdots \cdots}$

On adding above two equations, we get

$\red{\rm \: {e}^{x} + {e}^{ - x} = 2 + \frac{ 2{x}^{2} }{2!} + \frac{ 2{x}^{4} }{4!} + \frac{2 {x}^{6} }{6!} +\cdots \cdots}$

$\red{\rm \: {e}^{x} + {e}^{ - x} = 2\bigg(1 + \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} + \frac{{x}^{6} }{6!} +\cdots \cdots\bigg)}$

can be further rewritten as

$\red{\rm \: \dfrac{1}{2}({e}^{x} + {e}^{ - x}) = 1 + \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} + \frac{{x}^{6} }{6!} +\cdots \cdots}$

Now, on substituting x = $\frac{1}{2}$, we get

${\rm \: \dfrac{1}{2}({e}^{ \frac{1}{2} } + {e}^{ - \frac{1}{2} }) = 1 + \frac{ {( \frac{1}{2})}^{2} }{2!} + \frac{ {( \frac{1}{2})}^{4} }{4!} + \frac{{( \frac{1}{2})}^{6} }{6!} +\cdots \cdots}$

${\rm \: \dfrac{1}{2}\bigg( \sqrt{e} + \dfrac{1}{ \sqrt{e} } \bigg) = 1 + \frac{1}{4.2!} + \frac{1}{16.4!} + \frac{1}{64.6!} +\cdots \cdots}$

${\rm \: \dfrac{1}{2}\bigg(\dfrac{e + 1}{ \sqrt{e} } \bigg) = 1 + \frac{1}{4.2!} + \frac{1}{16.4!} + \frac{1}{64.6!} +\cdots \cdots}$

Hence,

$\rm\implies \boxed{ \rm{1+\dfrac{1}{4.2!}+\dfrac{1}{16.4!}+\dfrac{1}{64.6!} +\cdots \infty = \frac{e + 1}{2 \sqrt{e} } }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:{\rm \: e = 1 + 1+ \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots \cdots} }}$

$\boxed{ \rm{ \:{\rm \: {e}^{ - 1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + \cdots \cdots} }}$

$\boxed{ \rm{ \:{\rm \:e + {e}^{ - 1} =2\bigg(1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} +\cdots \cdots} \bigg)}}$

$\boxed{ \rm{ \:{\rm \:e - {e}^{ - 1} =2\bigg(1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} +\cdots \cdots} \bigg)}}$

Answer 2

Answer:

Given Question :- The sum of the series is

$\rm \: 1 + \dfrac{1}{4.2!} + \dfrac{1}{16.4!} + \dfrac{1}{64.6!} + \cdots \cdots \: \infty$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: 1 + \dfrac{1}{4.2!} + \dfrac{1}{16.4!} + \dfrac{1}{64.6!} + \cdots \cdots \: \infty$

We know, from exponential series

$\red{\rm \: {e}^{x} = 1 + x + \frac{ {x}^{2} }{2!} + \frac{ {x}^{3} }{3!} + \frac{ {x}^{4} }{4!} + \cdots \cdots}$

and

$\red{\rm \: {e}^{ - x} = 1 - x + \frac{ {x}^{2} }{2!} - \frac{ {x}^{3} }{3!} + \frac{ {x}^{4} }{4!} +\cdots \cdots}$

On adding above two equations, we get

$\red{\rm \: {e}^{x} + {e}^{ - x} = 2 + \frac{ 2{x}^{2} }{2!} + \frac{ 2{x}^{4} }{4!} + \frac{2 {x}^{6} }{6!} +\cdots \cdots}$

$\red{\rm \: {e}^{x} + {e}^{ - x} = 2\bigg(1 + \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} + \frac{{x}^{6} }{6!} +\cdots \cdots\bigg)}$

can be further rewritten as

$\red{\rm \: \dfrac{1}{2}({e}^{x} + {e}^{ - x}) = 1 + \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} + \frac{{x}^{6} }{6!} +\cdots \cdots}$

Now, on substituting x = $\frac{1}{2}$, we get

${\rm \: \dfrac{1}{2}({e}^{ \frac{1}{2} } + {e}^{ - \frac{1}{2} }) = 1 + \frac{ {( \frac{1}{2})}^{2} }{2!} + \frac{ {( \frac{1}{2})}^{4} }{4!} + \frac{{( \frac{1}{2})}^{6} }{6!} +\cdots \cdots}$

${\rm \: \dfrac{1}{2}\bigg( \sqrt{e} + \dfrac{1}{ \sqrt{e} } \bigg) = 1 + \frac{1}{4.2!} + \frac{1}{16.4!} + \frac{1}{64.6!} +\cdots \cdots}$

${\rm \: \dfrac{1}{2}\bigg(\dfrac{e + 1}{ \sqrt{e} } \bigg) = 1 + \frac{1}{4.2!} + \frac{1}{16.4!} + \frac{1}{64.6!} +\cdots \cdots}$

Hence,

$\rm\implies \boxed{ \rm{1+\dfrac{1}{4.2!}+\dfrac{1}{16.4!}+\dfrac{1}{64.6!} +\cdots \infty = \frac{e + 1}{2 \sqrt{e} } }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:{\rm \: e = 1 + 1+ \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots \cdots} }}$

$\boxed{ \rm{ \:{\rm \: {e}^{ - 1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + \cdots \cdots} }}$

$\boxed{ \rm{ \:{\rm \:e + {e}^{ - 1} =2\bigg(1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} +\cdots \cdots} \bigg)}}$

$\boxed{ \rm{ \:{\rm \:e - {e}^{ - 1} =2\bigg(1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} +\cdots \cdots} \bigg)}}$