Question

The sum of series $\frac{1}{2!}$+$\frac{1}{4!}$+$\frac{1}{6!}$+….. is –

• A. $\frac{(e^{2} - 1)}{2}$
• B. $\frac{(e - 1)^{2}}{2e}$
• C. $\frac{(e^{2} - 1)}{2e}$
• D. $\frac{(e^{2} - 2)}{e}$

Answer 1

Answer: (B)

$\frac{(e - 1)^{2}}{2e}$

Answer 2

We know that

e = 1 + $\frac{1}{1!}$+$\frac{1}{2!}$+$\frac{1}{3!}$+$\frac{1}{4!}$+…. …(i)

e^–1 = 1 – $\frac{1}{1!}$+$\frac{1}{2!}$–$\frac{1}{3!}$+$\frac{1}{4!}$–…. …(ii)

On adding Equation (i) and (ii), we get

e + e^–1 = 2 + $\frac{2}{2!}$ + $\frac{2}{4!}$+….

$\frac{e^{2} + 1}{e}$– 2 = $\frac{2}{2!}$+ $\frac{2}{4!}$+….

$\frac{e^{2} + 1 - 2e}{e}$= 2 $\left\lbrack \frac{1}{2!} + \frac{1}{4!} + ....\infty \right\rbrack$

$\frac{(e - 1)^{2}}{2e}$=$\frac{1}{2!}$+$\frac{1}{4!}$+…..