`1+(2x)/(1!)+(3x^(2))/(2!)+(4x^(3))/(3!)+..infty` is equal to

Options

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

Explanation

To solve the series \( S = 1 + \frac{2x}{1!} + \frac{3x^2}{2!} + \frac{4x^3}{3!} + \ldots \) up to infinity, we can follow these steps:

### Step 1: Identify the General Term
The series can be expressed in terms of its general term. The \( n \)-th term can be written as:
\[ T_n = \frac{(n+1)x^n}{n!} \]
This is because the first term corresponds to \( n=0 \), the second term corresponds to \( n=1 \), and so on.
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