Question

$(a^2-b^2)+\frac{a^4-b^4}{2!}+\frac{a^6-b^6}{3!}+...$ to $\infty$ is equal to

• A. $e^{a^2}-e^{b^2}$
• B. $e^{b^2}-e^{a^2}$
• C. $\frac{e^{a^2}}{e^{b^2}}$
• D. none of these

Answer 1

The correct option is A $e^{a^2}-e^{b^2}$

Simplifying, we get
$(a^2+\frac{a^4}{2!}+\frac{a^6}{3!}+...\infty )-(b^2+\frac{b^4}{2!}+\frac{b^6}{3!}+...\infty )$
$=(1+a^2+\frac{a^4}{2!}+\frac{a^6}{3!}+...\infty )-(1+b^2+\frac{b^4}{2!}+\frac{b^6}{3!}+...\infty )$
$=e^{a^2}-e^{b^2}$