If `S_(n)=(1^(2).(2))/(1!)+(2^(2).3)/(2!)+(3^(2).4)/(3!)+…(n^(2).(n+1))/(n!)` then
`lim_(n rarr infty) S_(n)` is equal to

Options

A. 3e
B. 5e
C. 7e
D. 9e

Answer

To solve the problem, we need to evaluate the limit of the series given by

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

as \( n \) approaches infinity.

### Step 1: Rewrite the Series

The general term of the series can be expressed as:

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

Thus, we can separate the series into two parts:

\[ S_n = \sum_{k=1}^{n} \frac{k^3}{k!} + \sum_{k=1}^{n} \frac{k^2}{k!} \]

### Step 2: Evaluate Each Series

We will evaluate the two series separately.

1. **For the first series**:

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

We can use the known result that:

\[ \sum_{k=0}^{\infty} \frac{k^3}{k!} = e \]

Therefore, as \( n \to \infty \):

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

2. **For the second series**:

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

Similarly, we have:

\[ \sum_{k=0}^{\infty} \frac{k^2}{k!} = e \]

Hence, as \( n \to \infty \):

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

### Step 3: Combine the Results

Now we can combine the results of both series:

\[ S_n \to e + e = 2e \]

### Step 4: Find the Limit

Finally, we find the limit as \( n \) approaches infinity:

\[ \lim_{n \to \infty} S_n = 2e \]

### Conclusion

Thus, the limit of \( S_n \) as \( n \) approaches infinity is:

\[ \lim_{n \to \infty} S_n = 2e \]