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Question: The sum of the series 1 + \(\frac{1}{4.2!}\)+\(\frac{1}{16.4!}\)+\(\frac{1}{64.6!}\)+…. is –...

The sum of the series 1 + 14.2!\frac{1}{4.2!}+116.4!\frac{1}{16.4!}+164.6!\frac{1}{64.6!}+…. is –

A

e+12e\frac{e + 1}{2\sqrt{e}}

B

e−12e\frac{e - 1}{2\sqrt{e}}

C

e+1e\frac{e + 1}{\sqrt{e}}

D

e–1e\frac{e–1}{\sqrt{e}}

Answer

e+12e\frac{e + 1}{2\sqrt{e}}

Explanation

Solution

We know that

ex = 1 + x +x22!\frac{x^{2}}{2!}+x33!\frac{x^{3}}{3!}+x44!\frac{x^{4}}{4!}+….

e–x = 1 – x +x22!\frac{x^{2}}{2!}–x33!\frac{x^{3}}{3!}+x44!\frac{x^{4}}{4!}–….

Ž ex+e–x2\frac{e^{x} + e^{–x}}{2} = 1 + x22!\frac{x^{2}}{2!}+ x44!\frac{x^{4}}{4!}+ x66!\frac{x^{6}}{6!}+….. Put, x = 12\frac{1}{2}

Ž e1/2+e–1/22\frac{e^{1/2} + e^{–1/2}}{2}= 1 + (12)2\left( \frac { 1 } { 2 } \right) ^ { 2 } 12!\frac{1}{2!}+ (12)4\left( \frac{1}{2} \right)^{4} 14!\frac{1}{4!}+….

Ž e+12e\frac{e + 1}{2\sqrt{e}} = 1 + (12)2\left( \frac { 1 } { 2 } \right) ^ { 2 } 12!\frac{1}{2!}+ (12)4\left( \frac{1}{2} \right)^{4} 14!\frac{1}{4!}+….