the value of `2/(1!)+ (2+4)/(2!) + (2+4+6)/(3!) +..................` is

Options

A. e
B. 2e
C. 3e
D. 3e - 2

Answer

To solve the problem, we need to evaluate the series:

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

### Step 1: Rewrite the series in a more manageable form

We can express the \(n\)-th term of the series as follows:

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

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

\[ \text{Sum} = n \cdot \left(\frac{a + l}{2}\right) = n \cdot \left(\frac{2 + 2n}{2}\right) = n(n + 1) \]

Thus, we can rewrite the \(n\)-th term as:

\[ \frac{n(n + 1)}{n!} \]

### Step 2: Simplify the expression

Now we can rewrite the series \(S\):

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

### Step 3: Break down the term

We can separate \(n(n + 1)\) as follows:

\[ n(n + 1) = n^2 + n \]

Thus, we can write:

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

### Step 4: Evaluate each sum

1. **For the sum \(\sum_{n=1}^{\infty} \frac{n}{n!}\)**:

We know that:

\[ \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \]

Differentiating both sides with respect to \(x\):

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

Setting \(x = 1\):

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

2. **For the sum \(\sum_{n=1}^{\infty} \frac{n^2}{n!}\)**:

We can use the result from the previous differentiation:

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

Differentiating again:

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

### Step 5: Combine the results

Now we can combine the results:

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

### Conclusion

Thus, the value of the series is:

\[ \boxed{3e} \]