Sum of series `(9)/(1!) + (19)/(2!) + (35)/(3!) + (57)/(4!) + (85)/(5!) + ...`

Options

A. `7e-3`
B. `12e-5`
C. `16e-5`
D. None

Answer

To solve the series \( S = \frac{9}{1!} + \frac{19}{2!} + \frac{35}{3!} + \frac{57}{4!} + \frac{85}{5!} + \ldots \), we will follow these steps:

### Step 1: Identify the Numerators
The numerators of the series are \( 9, 19, 35, 57, 85, \ldots \). We need to find a pattern in these numerators.

### Step 2: Find the Differences
Let's calculate the first differences of the numerators:
- \( 19 - 9 = 10 \)
- \( 35 - 19 = 16 \)
- \( 57 - 35 = 22 \)
- \( 85 - 57 = 28 \)

Now, the first differences are \( 10, 16, 22, 28 \).

### Step 3: Calculate the Second Differences
Next, we calculate the second differences:
- \( 16 - 10 = 6 \)
- \( 22 - 16 = 6 \)
- \( 28 - 22 = 6 \)

The second differences are constant, indicating that the numerators can be represented by a quadratic function.

### Step 4: Formulate the Quadratic Expression
Let’s assume the numerators can be expressed as \( a_n = An^2 + Bn + C \). We can use the first few terms to form equations:
1. For \( n=1 \): \( A(1^2) + B(1) + C = 9 \) → \( A + B + C = 9 \)
2. For \( n=2 \): \( A(2^2) + B(2) + C = 19 \) → \( 4A + 2B + C = 19 \)
3. For \( n=3 \): \( A(3^2) + B(3) + C = 35 \) → \( 9A + 3B + C = 35 \)

### Step 5: Solve the System of Equations
We can solve these equations step by step:
1. From equations 1 and 2, subtract the first from the second:
\[ (4A + 2B + C) - (A + B + C) = 19 - 9 \]
This simplifies to:
\[ 3A + B = 10 \quad \text{(Equation 4)} \]

2. Now, subtract equation 2 from equation 3:
\[ (9A + 3B + C) - (4A + 2B + C) = 35 - 19 \]
This simplifies to:
\[ 5A + B = 16 \quad \text{(Equation 5)} \]

3. Now we have a system of two equations (4) and (5):
\[ 3A + B = 10 \]
\[ 5A + B = 16 \]

4. Subtract equation 4 from equation 5:
\[ (5A + B) - (3A + B) = 16 - 10 \]
This gives:
\[ 2A = 6 \quad \Rightarrow \quad A = 3 \]

5. Substitute \( A = 3 \) back into equation 4:
\[ 3(3) + B = 10 \quad \Rightarrow \quad 9 + B = 10 \quad \Rightarrow \quad B = 1 \]

6. Substitute \( A = 3 \) and \( B = 1 \) back into equation 1:
\[ 3 + 1 + C = 9 \quad \Rightarrow \quad C = 5 \]

Thus, the formula for the numerators is:
\[ a_n = 3n^2 + n + 5 \]

### Step 6: Rewrite the Series
Now we can rewrite the series as:
\[ S = \sum_{n=1}^{\infty} \frac{3n^2 + n + 5}{n!} \]

### Step 7: Break Down the Series
We can separate the series:
\[ S = 3\sum_{n=1}^{\infty} \frac{n^2}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} + 5\sum_{n=1}^{\infty} \frac{1}{n!} \]

### Step 8: Evaluate Each Series
1. The series \( \sum_{n=1}^{\infty} \frac{n}{n!} = e \).
2. The series \( \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \).
3. The series \( \sum_{n=1}^{\infty} \frac{n^2}{n!} = e \) (can be derived using the identity \( n^2 = n(n-1) + n \)).

Thus, we have:
\[ S = 3e + e + 5(e - 1) = 3e + e + 5e - 5 = 12e - 5 \]

### Final Answer
The sum of the series is:
\[ S = 12e - 5 \]