`(1)/(1!) + (1 + 3)/(2!)*x + (1 + 3 + 5)/(3!)*x^(3) …`

Options

A. `e^(x) (1 + x)`
B. `e^(x) (1-x)`
C. `xe^(x)`
D. `e^(x) (x + 2)`

Answer

To solve the given series

\[ S = \frac{1}{1!} + \frac{1 + 3}{2!}x + \frac{1 + 3 + 5}{3!}x^3 + \ldots \]

we can start by identifying a pattern in the terms of the series.

### Step 1: Identify the general term

The series can be rewritten in terms of a summation. The \( n \)-th term of the series can be expressed as:

\[ \frac{1 + 3 + 5 + \ldots + (2n - 1)}{n!} x^{n-1} \]

The sum of the first \( n \) odd numbers \( 1 + 3 + 5 + \ldots + (2n - 1) \) is known to be \( n^2 \). Thus, we can rewrite the \( n \)-th term as:

\[ \frac{n^2}{n!} x^{n-1} \]

### Step 2: Rewrite the series

Now, we can express the entire series \( S \) as:

\[ S = \sum_{n=1}^{\infty} \frac{n^2}{n!} x^{n-1} \]

### Step 3: Factor out \( x \)

To simplify the summation, we can factor out \( x \):

\[ S = x \sum_{n=1}^{\infty} \frac{n^2}{n!} x^{n-2} \]

### Step 4: Use the known series expansion

We know that:

\[ \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \]

To find \( \sum_{n=1}^{\infty} \frac{n^2}{n!} x^{n-1} \), we can use the identity:

\[ \sum_{n=0}^{\infty} \frac{n^2 x^n}{n!} = x \frac{d}{dx} \left( e^x \right) = x e^x \]

Thus, we can write:

\[ \sum_{n=1}^{\infty} \frac{n^2 x^{n-1}}{n!} = e^x + x e^x \]

### Step 5: Substitute back into the equation

Now substituting this back into our expression for \( S \):

\[ S = x \left( e^x + x e^x \right) \]

### Step 6: Simplify the expression

This simplifies to:

\[ S = x e^x + x^2 e^x = e^x (x + x^2) \]

### Final Result

Thus, the final expression for the series is:

\[ S = e^x (1 + x) \]