Question

$1+\frac{a^2}{2!}+\frac{a^4}{4!}+\frac{a^6}{6!}+\dots \infty$

• A. $e^a+e^{-a}$
• B. $e^a-e^{-a}$
• C. $\frac{e^a+e^{-a}}{2}$
• D. $\frac{e^a-e^{-a}}{2}$

Answer 1

The correct option is C $\frac{e^a+e^{-a}}{2}$

$=\sum _{n=0}^{\infty }\frac{a^{2n}}{2n!}$
$=\frac{1}{2}\left(\sum _{n=0}^{a^n}\frac{a^n}{n!}+\sum _{n=0}^{\infty }\frac{(-a{)}^n}{n!}\right)$
$=\frac{1}{2}(e^a+e^{-a})$