The sum of the infinite series
`(2)/(1!) + (12)/(2!) + (28)/(3!) + (50)/(4!) + (78)/(5!) + ... = ..........`

Answer

To find the sum of the infinite series

\[ S = \frac{2}{1!} + \frac{12}{2!} + \frac{28}{3!} + \frac{50}{4!} + \frac{78}{5!} + \ldots \]

we first need to identify a pattern in the coefficients of the series. The coefficients are \(2, 12, 28, 50, 78, \ldots\).

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

Let's denote the coefficients as \(a_n\) where \(n\) starts from 1:

- \(a_1 = 2\)
- \(a_2 = 12\)
- \(a_3 = 28\)
- \(a_4 = 50\)
- \(a_5 = 78\)

Next, we can observe the differences between consecutive terms:

\[ \begin{align*} a_2 - a_1 & = 12 - 2 = 10 \\ a_3 - a_2 & = 28 - 12 = 16 \\ a_4 - a_3 & = 50 - 28 = 22 \\ a_5 - a_4 & = 78 - 50 = 28 \\ \end{align*} \]

The differences are \(10, 16, 22, 28\).

### Step 2: Find the second differences

Now, let's calculate the second differences:

\[ \begin{align*} 16 - 10 & = 6 \\ 22 - 16 & = 6 \\ 28 - 22 & = 6 \\ \end{align*} \]

The second differences are constant and equal to \(6\). This indicates that \(a_n\) is a quadratic function.

### Step 3: Assume a quadratic form

Assume \(a_n = An^2 + Bn + C\). We can use the first three terms to set up a system of equations:

1. For \(n=1\): \(A(1)^2 + B(1) + C = 2\) → \(A + B + C = 2\)
2. For \(n=2\): \(A(2)^2 + B(2) + C = 12\) → \(4A + 2B + C = 12\)
3. For \(n=3\): \(A(3)^2 + B(3) + C = 28\) → \(9A + 3B + C = 28\)

### Step 4: Solve the system of equations

Now we have the following system:

1. \(A + B + C = 2\) (1)
2. \(4A + 2B + C = 12\) (2)
3. \(9A + 3B + C = 28\) (3)

Subtract (1) from (2):

\[ (4A + 2B + C) - (A + B + C) = 12 - 2 \\ 3A + B = 10 \quad (4) \]

Subtract (2) from (3):

\[ (9A + 3B + C) - (4A + 2B + C) = 28 - 12 \\ 5A + B = 16 \quad (5) \]

Now, subtract (4) from (5):

\[ (5A + B) - (3A + B) = 16 - 10 \\ 2A = 6 \\ A = 3 \]

Substituting \(A = 3\) into (4):

\[ 3(3) + B = 10 \\ 9 + B = 10 \\ B = 1 \]

Now substituting \(A\) and \(B\) into (1):

\[ 3 + 1 + C = 2 \\ C = 2 - 4 = -2 \]

Thus, we have:

\[ a_n = 3n^2 + n - 2 \]

### Step 5: Write the series in terms of \(n\)

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

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

### Step 6: Split the series

We can split the series into three parts:

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

### Step 7: 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 \(\sum_{n=1}^{\infty} \frac{n^2}{n!} = e\) (using the identity \(\sum_{n=0}^{\infty} \frac{n^2}{n!} = e\)).

Now substituting these values back into \(S\):

\[ S = 3e + e - 2(e - 1) \\ = 3e + e - 2e + 2 \\ = 2e + 2 \]

### Final Answer

Thus, the sum of the infinite series is:

\[ \boxed{2e + 2} \]