The sum of the series is ` 9 + (16)/(2!) + (27)/(3!) + (42)/(4!) + ....oo`

Options

A. 5e
B. 7e
C. 9e
D. `11e-6`

Answer

To find the sum of the series \( S = 9 + \frac{16}{2!} + \frac{27}{3!} + \frac{42}{4!} + \ldots \), we will follow these steps:

### Step 1: Identify the general term of the series
The series can be expressed as:
\[ S = \sum_{n=1}^{\infty} \frac{T_n}{n!} \]
where \( T_n \) are the numerators of the series. Observing the numerators: \( 9, 16, 27, 42, \ldots \).

### Step 2: Find a pattern in the numerators
We can denote the numerators as:
- \( T_1 = 9 \)
- \( T_2 = 16 \)
- \( T_3 = 27 \)
- \( T_4 = 42 \)

To find a pattern, let's look at the differences between consecutive terms:
- \( T_2 - T_1 = 16 - 9 = 7 \)
- \( T_3 - T_2 = 27 - 16 = 11 \)
- \( T_4 - T_3 = 42 - 27 = 15 \)

The differences are \( 7, 11, 15 \), which themselves have a constant difference of \( 4 \). This indicates that \( T_n \) is a quadratic polynomial.

### Step 3: Assume a quadratic form for \( T_n \)
Let:
\[ T_n = an^2 + bn + c \]
Using the first few terms, we can set up equations:
1. For \( n=1 \): \( a(1^2) + b(1) + c = 9 \) → \( a + b + c = 9 \)
2. For \( n=2 \): \( a(2^2) + b(2) + c = 16 \) → \( 4a + 2b + c = 16 \)
3. For \( n=3 \): \( a(3^2) + b(3) + c = 27 \) → \( 9a + 3b + c = 27 \)

### Step 4: Solve the system of equations
From the equations:
1. \( a + b + c = 9 \) (1)
2. \( 4a + 2b + c = 16 \) (2)
3. \( 9a + 3b + c = 27 \) (3)

Subtract (1) from (2):
\[ (4a + 2b + c) - (a + b + c) = 16 - 9 \\ 3a + b = 7 \quad (4) \]

Subtract (2) from (3):
\[ (9a + 3b + c) - (4a + 2b + c) = 27 - 16 \\ 5a + b = 11 \quad (5) \]

Now, subtract (4) from (5):
\[ (5a + b) - (3a + b) = 11 - 7 \\ 2a = 4 \\ a = 2 \]

Substituting \( a = 2 \) into (4):
\[ 3(2) + b = 7 \\ 6 + b = 7 \\ b = 1 \]

Substituting \( a = 2 \) and \( b = 1 \) into (1):
\[ 2 + 1 + c = 9 \\ c = 6 \]

Thus, we have:
\[ T_n = 2n^2 + n + 6 \]

### Step 5: Substitute \( T_n \) back into the series
Now we can express the series \( S \):
\[ S = \sum_{n=1}^{\infty} \frac{2n^2 + n + 6}{n!} \]
This can be split into three separate sums:
\[ S = 2 \sum_{n=1}^{\infty} \frac{n^2}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} + 6 \sum_{n=1}^{\infty} \frac{1}{n!} \]

### Step 6: Evaluate each sum
1. \( \sum_{n=1}^{\infty} \frac{n}{n!} = e \) (from the series expansion of \( e^x \)).
2. \( \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \).
3. For \( \sum_{n=1}^{\infty} \frac{n^2}{n!} \), we can use the identity:
\[ \sum_{n=0}^{\infty} \frac{n^2}{n!} = e \]
Thus,
\[ \sum_{n=1}^{\infty} \frac{n^2}{n!} = e + 1 \]

### Step 7: Substitute back into \( S \)
Now substituting these results back into \( S \):
\[ S = 2(e + 1) + e + 6(e - 1) \]
\[ = 2e + 2 + e + 6e - 6 \]
\[ = 9e - 4 \]

### Conclusion
Thus, the sum of the series is:
\[ \boxed{9e - 4} \]