`4+(9)/(2!)+27/(3!)+....=`

Options

A. `e^2`
B. `e^3`
C. `e^4`
D. `1/(e^2)`

Answer

To solve the series \( 4 + \frac{9}{2!} + \frac{27}{3!} + \ldots \), we can recognize that the terms in the series can be expressed in a certain pattern. Let's break it down step by step.

### Step 1: Identify the Pattern
The series can be rewritten as:
\[ 4 + \frac{9}{2!} + \frac{27}{3!} + \ldots \]
Notice that:
- The first term is \( 4 = 3^1 + 1 \).
- The second term is \( \frac{9}{2!} = \frac{3^2}{2!} \).
- The third term is \( \frac{27}{3!} = \frac{3^3}{3!} \).

This suggests that the general term of the series can be expressed as:
\[ \frac{3^n}{n!} \quad \text{for } n \geq 1 \]

### Step 2: Write the Series in Summation Notation
We can express the series as:
\[ 4 + \sum_{n=1}^{\infty} \frac{3^n}{n!} \]

### Step 3: Recognize the Exponential Series
The series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \) for any real number \( x \). Therefore, we can write:
\[ \sum_{n=1}^{\infty} \frac{3^n}{n!} = e^3 - 1 \]

### Step 4: Combine the Results
Now substituting back into our series:
\[ 4 + \left( e^3 - 1 \right) = e^3 + 3 \]

### Step 5: Conclusion
Thus, the value of the series \( 4 + \frac{9}{2!} + \frac{27}{3!} + \ldots \) is:
\[ e^3 + 3 \]

### Final Answer
The final answer is \( e^3 + 3 \).