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^2y^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)

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