If `S_(n)=(1.2)/(3!)+(2.2^(2))/(4!)+(3.2^(2))/(5!)+...+` up to `n` terms, then sum of infinite terms is

Options

A. `(4)/(pi)`
B. `(3)/(e)`
C. `(pi)/(r )`
D. `1`

Explanation

To solve the problem, we need to find the sum of the infinite series given by:

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

### Step 1: Identify the General Term
The general term \( T_r \) of the series can be expressed as:

\[ T_r = \frac{r \cdot 2^r}{(r + 2)!} \]

This is because for \( r = 1 \), we have \( \frac{1 \cdot 2^1}{3!} \), for \( r = 2 \), we have \( \frac{2 \cdot 2^2}{4!} \), and so on.

**Hint:** To find the general term, look for a pattern in the series and express it in terms of \( r \).

### Step 2: Rewrite the General Term
We can rewrite \( T_r \) as follows:

\[ T_r = \frac{2^r}{(r + 2)!} \cdot r \]

This can also be expressed as:

\[ T_r = \frac{2^r}{(r + 2)!} - \frac{2^{r + 1}}{(r + 1)!} \]

This helps us to simplify the series.

**Hint:** Try to express the term in a form that can be summed easily.

### Step 3: Sum the Series
Now we can write \( S_n \) as:

\[ S_n = \sum_{r=1}^{n} \left( \frac{2^r}{(r + 2)!} - \frac{2^{r + 1}}{(r + 1)!} \right) \]

This can be simplified further by recognizing that it forms a telescoping series.

**Hint:** Look for cancellation in the series when summing.

### Step 4: Evaluate the Limit as \( n \to \infty \)
As \( n \) approaches infinity, we can evaluate the limit:

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

Taking the limit as \( n \to \infty \):

\[ S_{\infty} = \lim_{n \to \infty} \left( 1 - \frac{2^{n + 1}}{(n + 2)!} \right) \]

Since \( \frac{2^{n + 1}}{(n + 2)!} \) approaches 0 as \( n \) becomes very large, we find:

\[ S_{\infty} = 1 - 0 = 1 \]

**Hint:** Remember that factorial grows much faster than exponential functions.

### Final Answer
Thus, the sum of the infinite series is:

\[ S_{\infty} = 1 \]