The sum of the series
` (a + b ) (a - b) + (a + b)(a-b)(a^(2) + b^(2))/(2!) + `
` (a+ b) (a-b)(a^(4)+b^(4) +a^(2) b^(2))/(3!) + …"to" oo` lt is equal to

Options

A. (a)`e^(2^(2)) + e^(b^(2))`
B. (b)`e^(a^(2)+b^(2))`
C. (c)`e^(a^(2) -b^(2))`
D. (d)`e^(a^(2))-e^(b^(2))`

Answer

To find the sum of the series

\[ S = (a + b)(a - b) + \frac{(a + b)(a - b)(a^2 + b^2)}{2!} + \frac{(a + b)(a - b)(a^4 + b^4 + a^2 b^2)}{3!} + \ldots \]

we can break it down step by step.

### Step 1: Identify the first term
The first term of the series is:
\[ S_1 = (a + b)(a - b) = a^2 - b^2 \]

### Step 2: Identify the second term
The second term is:
\[ S_2 = \frac{(a + b)(a - b)(a^2 + b^2)}{2!} \]
Substituting \( S_1 \):
\[ S_2 = \frac{(a^2 - b^2)(a^2 + b^2)}{2!} = \frac{a^4 - b^4}{2!} \]

### Step 3: Identify the third term
The third term is:
\[ S_3 = \frac{(a + b)(a - b)(a^4 + b^4 + a^2 b^2)}{3!} \]
Using \( S_1 \):
\[ S_3 = \frac{(a^2 - b^2)(a^4 + b^4 + a^2 b^2)}{3!} \]
Expanding:
\[ S_3 = \frac{a^6 - b^6}{3!} \]

### Step 4: General term of the series
From the pattern observed, we can generalize the \( n \)-th term:
\[ S_n = \frac{(a^2 - b^2)(a^{2n} + b^{2n} + a^{2(n-1)}b^2 + \ldots)}{n!} \]
This can be simplified to:
\[ S_n = \frac{a^{2n} - b^{2n}}{n!} \]

### Step 5: Sum the series
Now we can sum the series:
\[ S = \sum_{n=1}^{\infty} \frac{a^{2n} - b^{2n}}{n!} \]
This can be separated into two series:
\[ S = \sum_{n=1}^{\infty} \frac{a^{2n}}{n!} - \sum_{n=1}^{\infty} \frac{b^{2n}}{n!} \]
Using the Taylor series expansion for \( e^x \):
\[ S = (e^{a^2} - 1) - (e^{b^2} - 1) \]
Thus:
\[ S = e^{a^2} - e^{b^2} \]

### Final Result
The sum of the series is:
\[ S = e^{a^2} - e^{b^2} \]