Question

If Sum of the series \(\frac{1}{1!}+\frac{1}{3!}+ \frac{1}{5!}+ ........\) = \(\frac{{{{\left( {1\ -\ e^2 } \right)}}}}{{xe}}\) then x will be equal to 12!+14!+16!+........ then 
1. 3
2. 2
3. 1
4. -2

Answer 1

Correct Answer - Option 4 : -2

Concept:

The expansion of ex is given as;

\({e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \infty \)  ----(1)

Calculation:

Put x = 1 in equation 1;

\(e = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + \ldots \infty \) ----(2)

Put x = -1 in equation 1;

\({e^{ - 1}} = 1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - \ldots \infty \) ----(3)

Substracting equation 2) and 3)

⇒ e - e^-1 = \(2(\frac{1}{1!}+\frac{1}{3!}+ \frac{1}{5!}+ ........)\)

⇒ \(e - \frac{1}{e} = 2\left[ {\frac{1}{1!}+\frac{1}{3!}+ \frac{1}{5!}+ ........ } \right]\)

⇒ \(\frac{e^2 - \ 1}{2e} = \left[ {\frac{1}{1!}+\frac{1}{3!}+ \frac{1}{5!}+ ........ } \right]\)

One comparing it from \(\frac{{{{\left( {1\ -\ e^2 } \right)}}}}{{xe}}\) = \(\frac{1}{1!}+\frac{1}{3!}+ \frac{1}{5!}+ ........\) we will get

x = -2