The sum of the series `((1)^(2).2)/(1!)+(2^(2).3)/(2!)+(3^(2).4)/(3!)+(4^(2).5)/(4!)`+..is

Options

A. 5e
B. 3e
C. 7e
D. 2e

Answer

To find the sum of the series

\[ S = \frac{1^2 \cdot 2}{1!} + \frac{2^2 \cdot 3}{2!} + \frac{3^2 \cdot 4}{3!} + \frac{4^2 \cdot 5}{4!} + \ldots \]

we can express this series in summation notation:

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

### Step 1: Simplifying the term inside the summation

The term inside the summation can be rewritten as:

\[ \frac{n^2 (n+1)}{n!} = \frac{n^3 + n^2}{n!} \]

Thus, we can split the summation into two parts:

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

### Step 2: Evaluating the first summation

For the first summation, we can use the known result:

\[ \sum_{n=0}^{\infty} \frac{n^k}{n!} = e \quad \text{for } k = 0, 1, 2, 3, \ldots \]

Specifically, we have:

\[ \sum_{n=1}^{\infty} \frac{n^3}{n!} = e \]

### Step 3: Evaluating the second summation

For the second summation, we can also use the known result:

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

### Step 4: Combining the results

Now, we can combine the results of both summations:

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

### Final Result

Thus, the sum of the series is:

\[ S = 2e \]