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Question: If \(T_{n} = \frac{3^{n}}{2(n!)} - \frac{1}{2(n!)},\) then \(S_{\infty} =\)...

If Tn=3n2(n!)−12(n!),T_{n} = \frac{3^{n}}{2(n!)} - \frac{1}{2(n!)}, then S∞=S_{\infty} =

A

e3−12\frac{e^{3} - 1}{2}

B

e3−e2\frac{e^{3} - e}{2}

C

e−32\frac{e - 3}{2}

D

None of these

Answer

e3−e2\frac{e^{3} - e}{2}

Explanation

Solution

Given that log⁡ex1+x\log_{e}\frac{x}{1 + x}

Therefore sum of the series

=11.2.3+13.4.5+15.6.7+.....∞=\frac{1}{1.2.3} + \frac{1}{3.4.5} + \frac{1}{5.6.7} + .....\infty =

=12[{1+31!+322!+….}−{1+11!+12!+….}]= \frac { 1 } { 2 } \left[ \left\{ 1 + \frac { 3 } { 1 ! } + \frac { 3 ^ { 2 } } { 2 ! } + \ldots .\right\} - \left\{ 1 + \frac { 1 } { 1 ! } + \frac { 1 } { 2 ! } + \ldots .\right\} \right] log⁡e2−12\log_{e}2 - \frac{1}{2}.