`(1.4)/(0!) + (2.5)/(1!) + (3.6)/(2!) + ... = `

Options

A. 11e
B. 5e
C. `e-4`
D. None

Answer

To solve the series \( S = \frac{1 \cdot 4}{0!} + \frac{2 \cdot 5}{1!} + \frac{3 \cdot 6}{2!} + \ldots \), we will derive a formula for the nth term and then sum the series.

### Step 1: Identify the nth term
The nth term of the series can be expressed as:
\[ T_n = \frac{n(n + 3)}{(n - 1)!} \]
This is because the first term corresponds to \( n = 1 \) giving \( \frac{1 \cdot 4}{0!} \), the second term corresponds to \( n = 2 \) giving \( \frac{2 \cdot 5}{1!} \), and so on.

### Step 2: Rewrite the series
We can rewrite the series \( S \) as:
\[ S = \sum_{n=1}^{\infty} \frac{n(n + 3)}{(n - 1)!} \]
This can be split into two separate sums:
\[ S = \sum_{n=1}^{\infty} \frac{n^2}{(n - 1)!} + 3 \sum_{n=1}^{\infty} \frac{n}{(n - 1)!} \]

### Step 3: Simplify the first sum
For the first sum:
\[ \sum_{n=1}^{\infty} \frac{n^2}{(n - 1)!} = \sum_{n=0}^{\infty} \frac{(n + 1)^2}{n!} = \sum_{n=0}^{\infty} \frac{n^2 + 2n + 1}{n!} \]
This can be separated into three sums:
\[ \sum_{n=0}^{\infty} \frac{n^2}{n!} + 2\sum_{n=0}^{\infty} \frac{n}{n!} + \sum_{n=0}^{\infty} \frac{1}{n!} \]

### Step 4: Evaluate the sums
Using the known series for \( e^x \):
- \( \sum_{n=0}^{\infty} \frac{1}{n!} = e \)
- \( \sum_{n=0}^{\infty} \frac{n}{n!} = e \)
- \( \sum_{n=0}^{\infty} \frac{n^2}{n!} = e \)

Thus:
\[ \sum_{n=0}^{\infty} \frac{n^2}{n!} = e + e = 2e \]
So, the first sum becomes:
\[ \sum_{n=1}^{\infty} \frac{n^2}{(n - 1)!} = 2e + 2e + e = 5e \]

### Step 5: Evaluate the second sum
For the second sum:
\[ 3 \sum_{n=1}^{\infty} \frac{n}{(n - 1)!} = 3 \cdot e \]

### Step 6: Combine the results
Now, combining both parts:
\[ S = 5e + 3e = 8e \]

### Final Result
Thus, the sum of the series is:
\[ S = 8e \]