Question

$\frac{1 + \frac{2^{2}}{2!} + \frac{2^{4}}{3!} + \frac{2^{6}}{4!} + .....\infty}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^{2}}{4!} + ....\infty} =$

• A. $e^{2}$
• B. $e^{2} - 1$
• C. $e^{3/2}$
• D. None of these

Answer 1

Answer: (B)

$e^{2} - 1$

Answer 2

$\log_{e}2 - 1$

$\log_{e}2$.