Question
The sum of series $$\frac{1}{{2!}} + \frac{1}{{4!}} + \frac{1}{{6!}} + ......{\text{ is}}$$
A.
$$\frac{{\left( {{e^2} - 2} \right)}}{e}$$
B.
$$\frac{{{{\left( {e - 1} \right)}^2}}}{{2e}}$$
C.
$$\frac{{\left( {{e^2} - 1} \right)}}{{2e}}$$
D.
$$\frac{{\left( {{e^2} - 1} \right)}}{2}$$
Answer :
$$\frac{{{{\left( {e - 1} \right)}^2}}}{{2e}}$$
Solution :
$$\eqalign{
& {\text{We know that }}e = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ...... \cr
& {\text{and }}{e^{ - 1}} = 1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ...... \cr
& \therefore \,\,e + {e^{ - 1}} = 2\left[ {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + ......} \right] \cr
& \therefore \,\,\frac{1}{{2!}} + \frac{1}{{4!}} + \frac{1}{{6!}} + ...... = \frac{{e + {e^{ - 1}}}}{2} - 1 \cr
& = \frac{{{e^2} + 1 - 2e}}{{2e}} \cr
& = \frac{{{{\left( {e - 1} \right)}^2}}}{{2e}} \cr} $$