The value of `(2)/(1!) + (2+4)/(2!) + (2+4+6)/(3!) +` … is

Options

A. e
B. 2e
C. 3e
D. None of these

Answer

To solve the problem, we need to evaluate the infinite series given by:

\[ S = \frac{2}{1!} + \frac{2 + 4}{2!} + \frac{2 + 4 + 6}{3!} + \ldots \]

### Step 1: Rewrite the series
We can express the series in a more manageable form. The \(n\)-th term of the series can be represented as:

\[ \frac{2 + 4 + 6 + \ldots + 2n}{n!} \]

The sum \(2 + 4 + 6 + \ldots + 2n\) can be simplified. This is an arithmetic series where the first term is 2, the last term is \(2n\), and the number of terms is \(n\). The sum of the first \(n\) even numbers is given by:

\[ 2 + 4 + 6 + \ldots + 2n = 2(1 + 2 + 3 + \ldots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]

Thus, we can rewrite the series \(S\) as:

\[ S = \sum_{n=1}^{\infty} \frac{n(n + 1)}{n!} \]

### Step 2: Split the series
Now, we can split the term \(n(n + 1)\) into two separate sums:

\[ S = \sum_{n=1}^{\infty} \frac{n^2}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} \]

### Step 3: Evaluate the second sum
The second sum can be evaluated as follows:

\[ \sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = e \]

### Step 4: Evaluate the first sum
For the first sum, we can use the identity \(n^2 = n(n-1) + n\):

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

The first part can be rewritten as:

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

Thus, we have:

\[ \sum_{n=1}^{\infty} \frac{n^2}{n!} = e + e = 2e \]

### Step 5: Combine the results
Now, substituting back into our expression for \(S\):

\[ S = 2e + e = 3e \]

### Final Answer
Thus, the value of the infinite series is:

\[ \boxed{3e} \]