Question

Sum of the series $\left(x+y\right)\left(x-y\right)+\frac{1}{2!}\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)+\frac{1}{3!}\left(x+y\right)\left(x-y\right)\left(x^4+y^4+x^2{y}^2\right)+\cdots$ 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

The correct option is D

$e^{x^2}-e^{y^2}$

Explanation for the correct option :

Formula to be used : We know that $\left(x+y\right)\left(x-y\right)=x^2-y^2$ and $\left(x-y\right)\left(x^2+y^2+xy\right)=x^3-y^3$.

Step-1 : Using the formula $\left(x+y\right)\left(x-y\right)=x^2-y^2$ in the given series

Using the above formula, the series becomes,

$$\begin{gathered} \left(x+y\right)\left(x-y\right)+\frac{1}{2!}\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)+\frac{1}{3!}\left(x+y\right)\left(x-y\right)\left(x^4+y^4+x^2{y}^2\right)+\cdots \\ =\left(x^2-y^2\right)+\frac{1}{2!}\left(x^2-y^2\right)\left(x^2+y^2\right)+\frac{1}{3!}\left(x^2-y^2\right)\left(x^4+y^4+x^2{y}^2\right)+\cdots \\ =\left(x^2-y^2\right)+\frac{1}{2!}\left(x^4-y^4\right)+\frac{1}{3!}\left(x^2-y^2\right)\left[{\left(x^2\right)}^2+x^2{y}^2+{\left(y^2\right)}^2\right]+\cdots \left[\because \left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x^4-y^4\right)\right] \end{gathered}$$

Step-2 : Using the formula $\left(x-y\right)\left(x^2+y^2+xy\right)=x^3-y^3$ in the above series

As $\left(x-y\right)\left(x^2+y^2+xy\right)=x^3-y^3$, $\left(x^2-y^2\right)\left[{\left(x^2\right)}^2+x^2{y}^2+{\left(y^2\right)}^2\right]=x^6-y^6$ and using this in the above series, we get

$$\begin{gathered} \left(x+y\right)\left(x-y\right)+\frac{1}{2!}\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)+\frac{1}{3!}\left(x+y\right)\left(x-y\right)\left(x^4+y^4+x^2{y}^2\right)+\cdots \\ =\left(x^2-y^2\right)+\frac{1}{2!}\left(x^4-y^4\right)+\frac{1}{3!}\left(x^2-y^2\right)\left[{\left(x^2\right)}^2+x^2{y}^2+{\left(y^2\right)}^2\right]+\cdots \\ =\left(x^2-y^2\right)+\frac{1}{2!}\left(x^4-y^4\right)+\frac{1}{3!}\left(x^6-y^6\right)+\cdots \end{gathered}$$

Step-3 : Getting a standard series

Formula to be used : We know that $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$

So, $1+x^2+\frac{{\left(x^2\right)}^2}{2!}+\frac{{\left(x^2\right)}^3}{3!}+\cdots =e^{x^2}$ and $1+y^2+\frac{{\left(y^2\right)}^2}{2!}+\frac{{\left(y^2\right)}^3}{3!}+\cdots =e^{y^2}$.

Using the above formula in the series obtained in the Step-2, we get

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

Hence, option (D) is the correct answer.