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

Options

A. e
B. `(3e)/2`
C. `(5e)/2`
D. `(7e)/2`

Answer

To solve the series

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

we can break it down step by step.

### Step 1: Rewrite the series

The series can be rewritten in terms of summation notation. The \(n\)-th term of the series can be expressed as:

\[ \frac{1 + 2 + 3 + \ldots + n}{n!} \]

Using the formula for the sum of the first \(n\) natural numbers, we have:

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

Thus, the \(n\)-th term becomes:

\[ \frac{\frac{n(n + 1)}{2}}{n!} = \frac{n(n + 1)}{2n!} \]

### Step 2: Simplify the terms

We can simplify the term further:

\[ \frac{n(n + 1)}{2n!} = \frac{1}{2} \cdot \frac{n(n + 1)}{n!} = \frac{1}{2} \cdot \frac{n + 1}{(n - 1)!} \]

Thus, the series can be rewritten as:

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

### Step 3: Split the series

We can split the series into two parts:

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

### Step 4: Evaluate the first sum

The first sum can be simplified:

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

### Step 5: Evaluate the second sum

The second sum can be simplified as follows:

\[ \sum_{n=2}^{\infty} \frac{1}{(n-2)!} = \sum_{m=0}^{\infty} \frac{1}{m!} = e \]

### Step 6: Combine results

Now, substituting back into the equation for \(S\):

\[ S = 1 + \frac{1}{2} \cdot e + \frac{1}{2} \cdot e = 1 + e \]

### Final Step: Conclusion

Thus, the final result for the series is:

\[ S = 1 + e \]