`1 + (2^(3))/(2!) + (3^(3))/(3!) + (4^(3))/(4!) +...oo=`

Options

A. 3e
B. 5e
C. 6e
D. 7e

Answer

To solve the series \( S = 1 + \frac{2^3}{2!} + \frac{3^3}{3!} + \frac{4^3}{4!} + \ldots \), we will follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:
\[ T_n = \frac{n^3}{n!} \]

### Step 2: Rewrite the General Term
We can rewrite \( T_n \) using the factorial definition:
\[ T_n = \frac{n^3}{n!} = \frac{n \cdot n^2}{n!} = \frac{n^2}{(n-1)!} \]
We can further simplify \( n^2 \) as:
\[ n^2 = n(n-1) + n \]
Thus, we can express \( T_n \) as:
\[ T_n = \frac{n(n-1)}{(n-2)!} + \frac{n}{(n-1)!} \]

### Step 3: Split the General Term
Now we can split \( T_n \) into two separate summations:
\[ T_n = \frac{n(n-1)}{(n-2)!} + \frac{n}{(n-1)!} \]

### Step 4: Set Up the Summation
Now we will sum \( T_n \) from \( n=1 \) to \( \infty \):
\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left( \frac{n(n-1)}{(n-2)!} + \frac{n}{(n-1)!} \right) \]

### Step 5: Adjust the Limits of Summation
We need to adjust the limits for each term:
1. For \( \sum_{n=1}^{\infty} \frac{n(n-1)}{(n-2)!} \), we can start from \( n=3 \):
\[ \sum_{n=3}^{\infty} \frac{n(n-1)}{(n-2)!} \]
This can be rewritten as:
\[ \sum_{k=1}^{\infty} \frac{k^2 + k}{k!} \]
where \( k = n-2 \).

2. For \( \sum_{n=1}^{\infty} \frac{n}{(n-1)!} \), we can start from \( n=2 \):
\[ \sum_{n=2}^{\infty} \frac{n}{(n-1)!} \]

### Step 6: Evaluate Each Summation
Using the known series expansions:
- The series \( \sum_{n=0}^{\infty} \frac{1}{n!} = e \)
- The series \( \sum_{n=1}^{\infty} \frac{n}{n!} = e \)
- The series \( \sum_{n=2}^{\infty} \frac{1}{(n-1)!} = e \)

Putting it all together:
\[ S = \sum_{n=3}^{\infty} \frac{n(n-1)}{(n-2)!} + \sum_{n=2}^{\infty} \frac{n}{(n-1)!} = 2e + e = 5e \]

### Final Result
Thus, the sum of the series is:
\[ S = 5e \]