The value of `(x+y)(x-y)+1/(2!)(x+y)(x-y)(x^2+y^2)+1/(3!)(x+y)(x-y)(x^4+y^4+x^2y^2)+`... is :

Options

A. `e^(x^(2))-e^(y^(2))`
B. `e^(x^(2))+e^(y^(2))`
C. `e^(x^(2))-(y^(2))`
D. `e^(x^(2))+(y^(2))`

Answer

To solve the given problem, we need to analyze the series step by step.

### Given Series:
\[ S = (x+y)(x-y) + \frac{1}{2!}(x+y)(x-y)(x^2+y^2) + \frac{1}{3!}(x+y)(x-y)(x^4+y^4+x^2y^2) + \ldots \]

### Step 1: Factor Out Common Terms
Notice that each term in the series has a common factor of \((x+y)(x-y)\). We can factor this out:
\[ S = (x+y)(x-y) \left( 1 + \frac{1}{2!}(x^2+y^2) + \frac{1}{3!}(x^4+y^4+x^2y^2) + \ldots \right) \]

### Step 2: Simplify the Series Inside the Parentheses
Now, let's denote the series inside the parentheses as \(T\):
\[ T = 1 + \frac{1}{2!}(x^2+y^2) + \frac{1}{3!}(x^4+y^4+x^2y^2) + \ldots \]

### Step 3: Identify the Pattern in the Series
The terms in \(T\) can be recognized as the series expansion of \(e^{x^2}\) and \(e^{y^2}\):
- The term \(1\) corresponds to \(e^0\).
- The term \(\frac{x^2}{1!}\) corresponds to the first term of the series for \(e^{x^2}\).
- The term \(\frac{x^4}{2!}\) corresponds to the second term of the series for \(e^{x^2}\), and so on.

Thus, we can express \(T\) in terms of exponential functions:
\[ T = e^{x^2} - 1 + e^{y^2} - 1 \]

### Step 4: Combine the Results
Now we can substitute \(T\) back into our expression for \(S\):
\[ S = (x+y)(x-y) \left( e^{x^2} - 1 + e^{y^2} - 1 \right) \]
\[ = (x+y)(x-y) \left( e^{x^2} + e^{y^2} - 2 \right) \]

### Step 5: Final Result
Thus, the value of the original series is:
\[ S = (x+y)(x-y) \left( e^{x^2} - e^{y^2} \right) \]