Question

The value of $(a+b)(a-b)+\frac{1}{2!}(a+b)(a-b)(a^2+b^2)+\frac{1}{3!}(a+b)(a-b)(a^4+a^2{b}^2+b^4)+...$ is

• A. $e^{a^2}-e^{b^2}$
• B. $e^{a^2}+e^{b^2}$
• C. $e^{a^2-b^2}$
• D. None of these

Answer 1

Answer: (A)

$e^{a^2}-e^{b^2}$

Upon simplifying the above expression, we get
$(a^2-b^2)+\frac{a^4-b^4}{2!}+\frac{a^6-b^6}{3!}...\infty$
$=[a^2+\frac{(a^2{)}^2}{2!}+\frac{(a^2{)}^3}{3!}+...]-[b^2+\frac{(b^2{)}^2}{2!}+\frac{(b^2{)}^3}{3!}+...]$
$=[e^{a^2}-1]-[e^{b^2}-1]$
$=e^{a^2}-1-e^{b^2}+1$
$=e^{a^2}-e^{b^2}$