The sum of the series
`1+(1^2+2^2)/(2!)+(1^(2)+2^(2)+3^(2))/(3!)+(1^(2)+2^(2)+3^(2)+4^2)/(4!)`+.. Is

Options

A. 3e
B. `(17)/(6)e`
C. `(31)/(6)e`
D. `(13)/(6)e`

Answer

To find the sum of the series

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

we can break it down step by step.

### Step 1: Identify the General Term
The general 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
The sum of the first \( n \) squares is given by the formula:

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

Thus, we can rewrite \( T_n \):

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

### Step 3: Write the Series
Now, we can express 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 Series
We can factor out the constant \( \frac{1}{6} \):

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

### Step 5: Break Down the Series
We can express \( n(n + 1)(2n + 1) \) in terms of factorials:

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

Thus, we can separate the series:

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

### Step 6: Evaluate Each Series
Using the known series expansions:

1. \( \sum_{n=0}^{\infty} \frac{n}{n!} = e \)
2. \( \sum_{n=0}^{\infty} \frac{n^2}{n!} = e \) (using the fact that \( n^2 = n + n(n-1) \))
3. \( \sum_{n=0}^{\infty} \frac{n^3}{n!} = e \) (similar reasoning)

Thus:

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

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

\[ S = \frac{1}{6} \left( 2(3e - 2) + 3e + (e - 1) \right) \]

This simplifies to:

\[ S = \frac{1}{6} \left( 6e - 4 + 3e + e - 1 \right) = \frac{1}{6} \left( 10e - 5 \right) = \frac{5e - \frac{5}{2}}{3} \]

### Final Result
Thus, the sum of the series is:

\[ S = \frac{5e - \frac{5}{2}}{3} \]