The sum of `(2)/(1!) + (6)/(2!) + (12)/(3!) + (20)/(4!) + ...` is

Options

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

Answer

To find the sum of the series \( S = \frac{2}{1!} + \frac{6}{2!} + \frac{12}{3!} + \frac{20}{4!} + \ldots \), we will first identify a pattern in the numerators.

### Step 1: Identify the Numerators
The numerators of the series are \( 2, 6, 12, 20, \ldots \). Let's denote the \( n \)-th term of the numerators as \( T_n \).

Calculating the first few terms:
- \( T_1 = 2 \)
- \( T_2 = 6 \)
- \( T_3 = 12 \)
- \( T_4 = 20 \)

Now, let's examine the differences between consecutive terms:
- \( T_2 - T_1 = 6 - 2 = 4 \)
- \( T_3 - T_2 = 12 - 6 = 6 \)
- \( T_4 - T_3 = 20 - 12 = 8 \)

The differences are \( 4, 6, 8 \), which are increasing by \( 2 \). This suggests that the numerators can be expressed in a quadratic form.

### Step 2: Find a Formula for the Numerators
Assuming \( T_n = an^2 + bn + c \), we can set up equations based on the known values:
1. \( T_1 = a(1^2) + b(1) + c = 2 \)
2. \( T_2 = a(2^2) + b(2) + c = 6 \)
3. \( T_3 = a(3^2) + b(3) + c = 12 \)

This gives us the following system of equations:
1. \( a + b + c = 2 \)
2. \( 4a + 2b + c = 6 \)
3. \( 9a + 3b + c = 12 \)

Subtracting the first equation from the second:
\[ (4a + 2b + c) - (a + b + c) = 6 - 2 \implies 3a + b = 4 \quad \text{(Equation 4)} \]

Subtracting the second equation from the third:
\[ (9a + 3b + c) - (4a + 2b + c) = 12 - 6 \implies 5a + b = 6 \quad \text{(Equation 5)} \]

Now we can subtract Equation 4 from Equation 5:
\[ (5a + b) - (3a + b) = 6 - 4 \implies 2a = 2 \implies a = 1 \]

Substituting \( a = 1 \) into Equation 4:
\[ 3(1) + b = 4 \implies b = 1 \]

Now substituting \( a \) and \( b \) back into the first equation:
\[ 1 + 1 + c = 2 \implies c = 0 \]

Thus, the formula for the numerators is:
\[ T_n = n^2 + n = n(n + 1) \]

### Step 3: Rewrite the Series
Now we can rewrite the series \( S \):
\[ S = \sum_{n=1}^{\infty} \frac{T_n}{n!} = \sum_{n=1}^{\infty} \frac{n(n + 1)}{n!} \]

This can be split into two parts:
\[ S = \sum_{n=1}^{\infty} \frac{n^2}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} \]

### Step 4: Evaluate Each Part
1. For \( \sum_{n=1}^{\infty} \frac{n}{n!} \):
\[ \sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = e \]

2. For \( \sum_{n=1}^{\infty} \frac{n^2}{n!} \):
\[ \sum_{n=1}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + e = e + e = 2e \]

### Step 5: Combine the Results
Combining both parts:
\[ S = 2e + e = 3e \]

### Final Answer
The sum of the series is:
\[ \boxed{3e} \]