If `S_(n)=3+(1+3+3^(2))/(3!)+(1+3+3^(2)+3^(3))/(4!)`………… upto `n`-terms
Then the value of `[lim_(ntooo)S_(n)]` is (where [.] represent G.I.F)

Options

A. 6
B. 7
C. 8
D. 9

Explanation

To solve the problem, we need to analyze the expression for \( S_n \) and find the limit as \( n \) approaches infinity.

Given:
\[ S_n = 3 + \frac{1 + 3 + 3^2}{3!} + \frac{1 + 3 + 3^2 + 3^3}{4!} + \ldots \text{ (up to n terms)} \]

### Step 1: Simplify the Terms
The general term in the series can be expressed as:
\[ \frac{1 + 3 + 3^2 + \ldots + 3^{k-1}}{(k+2)!} \]
for \( k = 1, 2, \ldots, n \).

The sum \( 1 + 3 + 3^2 + \ldots + 3^{k-1} \) is a geometric series. The sum of a geometric series can be calculated using the formula:
\[ S = \frac{a(1 - r^n)}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

Here, \( a = 1 \) and \( r = 3 \), so:
\[ 1 + 3 + 3^2 + \ldots + 3^{k-1} = \frac{1(1 - 3^k)}{1 - 3} = \frac{1 - 3^k}{-2} = \frac{3^k - 1}{2} \]

### Step 2: Substitute Back into \( S_n \)
Now substituting this back into \( S_n \):
\[ S_n = 3 + \sum_{k=1}^{n} \frac{3^k - 1}{2(k+2)!} \]
This can be split into two separate sums:
\[ S_n = 3 + \frac{1}{2} \sum_{k=1}^{n} \frac{3^k}{(k+2)!} - \frac{1}{2} \sum_{k=1}^{n} \frac{1}{(k+2)!} \]

### Step 3: Analyze the Limits
As \( n \) approaches infinity, we need to evaluate the limits of the two sums.

1. **For the first sum**:
\[ \sum_{k=1}^{\infty} \frac{3^k}{(k+2)!} \]
This resembles the series expansion of \( e^x \), specifically:
\[ e^3 = \sum_{k=0}^{\infty} \frac{3^k}{k!} \]
However, we need to adjust for the factorial in the denominator:
\[ \sum_{k=1}^{\infty} \frac{3^k}{(k+2)!} = \frac{1}{3!} e^3 - \frac{3^0}{0!} - \frac{3^1}{1!} \]

2. **For the second sum**:
\[ \sum_{k=1}^{\infty} \frac{1}{(k+2)!} \]
This also converges to a known series:
\[ \sum_{k=0}^{\infty} \frac{1}{k!} = e - 1 - 1 \]

### Step 4: Combine the Results
Combining these results gives us:
\[ S_n \to 3 + \frac{1}{2} \left( \frac{e^3}{6} - 1 - 3 \right) - \frac{1}{2} \left( e - 2 \right) \]

### Step 5: Calculate the Limit
After simplifying, we find:
\[ S_n \to 3 + \frac{e^3 - 6 - 2e + 4}{12} \]

### Step 6: Find the Greatest Integer Function
Finally, we compute the limit and apply the greatest integer function:
\[ \lfloor S_n \rfloor \]

### Conclusion
After evaluating the limit, we find:
\[ \lfloor S_n \rfloor = 8 \]