Sum of infinity the series
`1 + (1^(2) + 2^(2) )/(2!) + (1^(2) + 2^(2) + 3^(2))/(3!) + ...=`

Options

A. 11e
B. 13e
C. `(17)/(6)e`
D. `(5)/(3)e`

Answer

To find the sum of the infinite series

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

we will derive a formula for the \(n\)-th term and then sum the series.

### Step 1: Identify the \(n\)-th term

The \(n\)-th term of the series can be expressed as:

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

### Step 2: Use the formula for the sum of squares

We know that the sum of the squares of the first \(n\) natural numbers is given by:

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

Substituting this into our expression for \(T_n\):

\[ T_n = \frac{\frac{n(n + 1)(2n + 1)}{6}}{n!} = \frac{n(n + 1)(2n + 1)}{6n!} \]

### Step 3: Rewrite the series

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

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{n(n + 1)(2n + 1)}{6n!} \]

### Step 4: Simplify the expression

We can factor out \(\frac{1}{6}\):

\[ S = \frac{1}{6} \sum_{n=1}^{\infty} \frac{n(n + 1)(2n + 1)}{n!} \]

### Step 5: Break down the summation

We can break down the summation into simpler parts. Notice that:

\[ n(n + 1)(2n + 1) = 2n^3 + 3n^2 + n \]

Thus, we can write:

\[ S = \frac{1}{6} \left( 2 \sum_{n=1}^{\infty} \frac{n^3}{n!} + 3 \sum_{n=1}^{\infty} \frac{n^2}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} \right) \]

### Step 6: Evaluate the individual summations

Using the known results for the sums involving \(n\):

1. \(\sum_{n=0}^{\infty} \frac{n}{n!} = e\)
2. \(\sum_{n=0}^{\infty} \frac{n^2}{n!} = e\)
3. \(\sum_{n=0}^{\infty} \frac{n^3}{n!} = e\)

We can adjust these sums for \(n\) starting from 1:

- \(\sum_{n=1}^{\infty} \frac{n}{n!} = e\)
- \(\sum_{n=1}^{\infty} \frac{n^2}{n!} = e + e = 2e\)
- \(\sum_{n=1}^{\infty} \frac{n^3}{n!} = e + 3e + e = 5e\)

### Step 7: Substitute back into the expression for \(S\)

Now substituting these back into our expression for \(S\):

\[ S = \frac{1}{6} \left( 2 \cdot 5e + 3 \cdot 2e + 1 \cdot e \right) = \frac{1}{6} (10e + 6e + e) = \frac{1}{6} (17e) \]

### Final Result

Thus, the sum of the infinite series is:

\[ S = \frac{17e}{6} \]