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Question: \[1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + ....\infty =\]...

1+31!+52!+73!+....∞=1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + ....\infty =

A

ee

B

2e2e

C

3e3e

D

4e4e

Answer

3e3e

Explanation

Solution

log⁡e(n2+1n2)\log_{e}\left( \frac{n^{2} + 1}{n^{2}} \right)

log⁡e(n2n2−1)\log_{e}\left( \frac{n^{2}}{n^{2} - 1} \right)

m−nm+n+13(m−nm+n)3+15(m−nm+n)5+......∞=\frac{m - n}{m + n} + \frac{1}{3}\left( \frac{m - n}{m + n} \right)^{3} + \frac{1}{5}\left( \frac{m - n}{m + n} \right)^{5} + ......\infty =.