The sum of the series `(1^(2).2^(2))/(1!)+(2^(2).3^(2))/(2!)+(3^(2).4^(2))/(3!)`+.. Is

Options

A. 27e
B. 24e
C. 28e
D. 25e

Answer

To find the sum of the series

\[ S = \sum_{n=1}^{\infty} \frac{n^2 (n+1)^2}{n!} \]

we can break it down step by step.

### Step 1: Rewrite the Series
The series can be rewritten as:

\[ S = \sum_{n=1}^{\infty} \frac{n^2 (n^2 + 2n + 1)}{n!} \]

This expands to:

\[ S = \sum_{n=1}^{\infty} \left( \frac{n^4}{n!} + \frac{2n^3}{n!} + \frac{n^2}{n!} \right) \]

### Step 2: Separate the Series
Now, we can separate the series into three parts:

\[ S = \sum_{n=1}^{\infty} \frac{n^4}{n!} + 2 \sum_{n=1}^{\infty} \frac{n^3}{n!} + \sum_{n=1}^{\infty} \frac{n^2}{n!} \]

### Step 3: Evaluate Each Series
We will evaluate each of these series using known results.

1. **For** \( \sum_{n=0}^{\infty} \frac{n^2}{n!} \):
- It is known that \( \sum_{n=0}^{\infty} \frac{n^k}{n!} = e \) for \( k = 0, 1, 2, \ldots \)
- Specifically, \( \sum_{n=0}^{\infty} \frac{n^2}{n!} = e \).

2. **For** \( \sum_{n=0}^{\infty} \frac{n^3}{n!} \):
- It can be shown that \( \sum_{n=0}^{\infty} \frac{n^3}{n!} = e \cdot (1 + 1 + 1) = 3e \).

3. **For** \( \sum_{n=0}^{\infty} \frac{n^4}{n!} \):
- Similarly, \( \sum_{n=0}^{\infty} \frac{n^4}{n!} = e \cdot (1 + 1 + 1 + 1) = 7e \).

### Step 4: Combine the Results
Now we can substitute back into our expression for \( S \):

\[ S = 7e + 2(6e) + e = 7e + 12e + e = 20e \]

### Step 5: Final Result
Thus, the sum of the series is:

\[ S = 20e \]