Question

Sum of the series
$(x+y)(x-y)+\frac{1}{2!}(x+y)(x-y)(x^2+y^2)$
+ $\frac{1}{3!}(x+y)(x-y)(x^4+y^4+x^2{y}^2)+....\infty$ is

• A. $e^x+e^y$
• B. $e^x-e^y$
• C. $e^{x^2}+e^{y^2}$
• D. $e^{x^2}-e^{y^2}$

Answer 1

Answer: (D)

$\left(x+y\right)\left(x-y\right)+\frac{1}{2!}\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)$
$+\frac{1}{3!}\left(x+y\right)\left(x-y\right)\left(x^4+y^4+x^2{y}^2\right)+\dots$
$=\left(x^2-y^2\right)+\frac{1}{2!}\left(x^4-y^4\right)+\frac{1}{3!}\left(x^2-y^2\right)$
$\left(x^4+y^4+x^2{y}^2\right)$
$=\left(x^2-y^2\right)+\frac{1}{2!}\left(x^4-y^4\right)+\frac{1}{3!}\left(x^6-y^6\right)+\dots$
$=\left(x^2+\frac{1}{2!}{x}^4+\frac{x^6}{3!}+\dots \right)-\left(y^2+\frac{y^4}{2!}+\frac{y^6}{3!}+\dots \right)$
$=\left(1+\frac{x^2}{1!}+\frac{{\left(x^2\right)}^2}{2!}+\frac{{\left(x^2\right)}^3}{3!}+\dots \right)$
$-\left(1+\frac{y^2}{1!}+\frac{{\left(y^2\right)}^2}{2!}+\frac{{\left(y^2\right)}^3}{3!}+\dots \right)$
$=e^{x^2}-e^{y^2}$