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

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

we can break down the general term and evaluate it step by step.

### Step 1: Write the general term
The general term of the series can be expressed as:

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

### Step 2: Separate the series
We can separate the series into two parts:

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

### Step 3: Evaluate the first series \(\sum_{n=1}^{\infty} \frac{n^3}{n!}\)
To evaluate \(\sum_{n=1}^{\infty} \frac{n^3}{n!}\), we can use the identity:

\[ n^3 = n(n-1)(n-2) + 3n(n-1) + n \]

Thus,

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

Now, we can simplify each part:

1. \(\sum_{n=1}^{\infty} \frac{n(n-1)(n-2)}{n!} = \sum_{n=3}^{\infty} \frac{1}{(n-3)!} = e\)
2. \(\sum_{n=1}^{\infty} \frac{n(n-1)}{n!} = \sum_{n=2}^{\infty} \frac{1}{(n-2)!} = e\)
3. \(\sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = e\)

Putting it all together:

\[ \sum_{n=1}^{\infty} \frac{n^3}{n!} = e + 3e + e = 5e \]

### Step 4: Evaluate the second series \(\sum_{n=1}^{\infty} \frac{n^2}{n!}\)
Using a similar approach:

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

1. \(\sum_{n=1}^{\infty} \frac{n(n-1)}{n!} = e\)
2. \(\sum_{n=1}^{\infty} \frac{n}{n!} = e\)

Thus,

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

### Step 5: Combine the results
Now we can combine the results from both series:

\[ S = 5e + 2e = 7e \]

### Final Answer
The sum of the series is

\[ \boxed{7e} \]