The sum of the series `(12)/(2!)+(28)/(3!)+(50)/(4!)+(78)/(5!)+`…is

Options

A. e
B. 3e
C. 4e
D. 5e

Answer

To find the sum of the series \(\frac{12}{2!} + \frac{28}{3!} + \frac{50}{4!} + \frac{78}{5!} + \ldots\), we will first identify the pattern in the numerators and then express the series in a more manageable form.

### Step 1: Identify the pattern in the numerators

The numerators of the series are:
- \(12\)
- \(28\)
- \(50\)
- \(78\)

Let's analyze the differences between consecutive terms:
- \(28 - 12 = 16\)
- \(50 - 28 = 22\)
- \(78 - 50 = 28\)

Now, let's analyze the differences of these differences:
- \(22 - 16 = 6\)
- \(28 - 22 = 6\)

Since the second differences are constant, the numerators form a quadratic sequence. We can express the \(n\)-th term of the numerators as a quadratic function.

### Step 2: Find the quadratic function

Assume the \(n\)-th term can be expressed as:
\[ T_n = an^2 + bn + c \]
We can set up equations using the known values:
1. For \(n=1\): \(T_1 = 12 \Rightarrow a(1^2) + b(1) + c = 12\)
2. For \(n=2\): \(T_2 = 28 \Rightarrow a(2^2) + b(2) + c = 28\)
3. For \(n=3\): \(T_3 = 50 \Rightarrow a(3^2) + b(3) + c = 50\)

This gives us the system of equations:
1. \(a + b + c = 12\)
2. \(4a + 2b + c = 28\)
3. \(9a + 3b + c = 50\)

### Step 3: Solve the system of equations

Subtract the first equation from the second:
\[ (4a + 2b + c) - (a + b + c) = 28 - 12 \]
\[ 3a + b = 16 \quad \text{(Equation 4)} \]

Subtract the second equation from the third:
\[ (9a + 3b + c) - (4a + 2b + c) = 50 - 28 \]
\[ 5a + b = 22 \quad \text{(Equation 5)} \]

Now, subtract Equation 4 from Equation 5:
\[ (5a + b) - (3a + b) = 22 - 16 \]
\[ 2a = 6 \Rightarrow a = 3 \]

Substituting \(a = 3\) back into Equation 4:
\[ 3(3) + b = 16 \Rightarrow 9 + b = 16 \Rightarrow b = 7 \]

Now substitute \(a\) and \(b\) into the first equation:
\[ 3 + 7 + c = 12 \Rightarrow c = 2 \]

Thus, the \(n\)-th term of the numerators is:
\[ T_n = 3n^2 + 7n + 2 \]

### Step 4: Write the series in summation form

The series can now be expressed as:
\[ S = \sum_{n=1}^{\infty} \frac{3n^2 + 7n + 2}{n!} \]

### Step 5: Split the series

We can split the series into three separate series:
\[ S = 3 \sum_{n=1}^{\infty} \frac{n^2}{n!} + 7 \sum_{n=1}^{\infty} \frac{n}{n!} + 2 \sum_{n=1}^{\infty} \frac{1}{n!} \]

### Step 6: Evaluate the series

Using known results:
1. \(\sum_{n=0}^{\infty} \frac{1}{n!} = e\)
2. \(\sum_{n=1}^{\infty} \frac{n}{n!} = e\)
3. \(\sum_{n=1}^{\infty} \frac{n^2}{n!} = e + e = 2e\)

Now substituting these results into our expression for \(S\):
\[ S = 3(2e) + 7(e) + 2(e) = 6e + 7e + 2e = 15e \]

### Final Answer:
The sum of the series is \(15e\).