`1 +(2^3)/(1!) x + (3^3)/(2!) x^2 + ....oo` =

Options

A. `(x^3+6x^(2) -7x -1)e^(x)`
B. `(x^3+6x^(2) +7x +1)e^(x)`
C. `(x^3+7x^(2) -6x +1)e^(x)`
D. `(x^3+x^(2) -2x -1)e^(x)`

Answer

To solve the series \( S = 1 + \frac{2^3}{1!} x + \frac{3^3}{2!} x^2 + \ldots \), we need to identify the general term and express the series in a more manageable form.

### Step-by-Step Solution:

1. **Identify the General Term**:
The series can be expressed as:
\[ S = \sum_{n=0}^{\infty} \frac{(n+1)^3}{n!} x^n \]
Here, the general term is \( \frac{(n+1)^3}{n!} x^n \).

2. **Expand the Cubic Term**:
We can expand \( (n+1)^3 \) using the binomial theorem:
\[ (n+1)^3 = n^3 + 3n^2 + 3n + 1 \]
Thus, we can rewrite the series:
\[ S = \sum_{n=0}^{\infty} \frac{n^3}{n!} x^n + 3\sum_{n=0}^{\infty} \frac{n^2}{n!} x^n + 3\sum_{n=0}^{\infty} \frac{n}{n!} x^n + \sum_{n=0}^{\infty} \frac{1}{n!} x^n \]

3. **Recognize Standard Series**:
The series \( \sum_{n=0}^{\infty} \frac{1}{n!} x^n \) is the Taylor series for \( e^x \). We will also need the derivatives of this series:
- \( \sum_{n=0}^{\infty} \frac{n}{n!} x^n = x e^x \)
- \( \sum_{n=0}^{\infty} \frac{n^2}{n!} x^n = x \frac{d}{dx}(e^x) = x e^x + e^x = (x+1)e^x \)
- \( \sum_{n=0}^{\infty} \frac{n^3}{n!} x^n = x \frac{d}{dx}((x+1)e^x) = x(e^x + (x+1)e^x) = (x^2 + 3x + 1)e^x \)

4. **Combine the Results**:
Now substituting these results back into our expression for \( S \):
\[ S = (x^2 + 3x + 1)e^x + 3(x + 1)e^x + 3xe^x + e^x \]
Simplifying this gives:
\[ S = (x^2 + 3x + 1 + 3x + 3 + 3x + 1)e^x = (x^2 + 9x + 5)e^x \]

5. **Final Result**:
Thus, the final result for the series is:
\[ S = (x^2 + 9x + 5)e^x \]