Sum to n terms the series
`4 + 14 + 30 + 52+ 82+ 114+.........`

Answer

To find the sum to n terms of the series \(4 + 14 + 30 + 52 + 82 + 114 + \ldots\), we will follow these steps:

### Step 1: Identify the pattern in the series
First, we observe the given series:
- The first term \(a_1 = 4\)
- The second term \(a_2 = 14\)
- The third term \(a_3 = 30\)
- The fourth term \(a_4 = 52\)
- The fifth term \(a_5 = 82\)
- The sixth term \(a_6 = 114\)

Now, let's find the differences between consecutive terms:
- \(a_2 - a_1 = 14 - 4 = 10\)
- \(a_3 - a_2 = 30 - 14 = 16\)
- \(a_4 - a_3 = 52 - 30 = 22\)
- \(a_5 - a_4 = 82 - 52 = 30\)
- \(a_6 - a_5 = 114 - 82 = 32\)

Next, we find the second differences:
- \(16 - 10 = 6\)
- \(22 - 16 = 6\)
- \(30 - 22 = 8\)
- \(32 - 30 = 2\)

The second differences are not constant, indicating that the terms are not in a simple arithmetic progression. However, we can see that the first differences are increasing.

### Step 2: Establish a general formula for the nth term
We can observe that the nth term appears to be a quadratic function. Let's assume:
\[ a_n = An^2 + Bn + C \]
We will use the first few terms to set up equations to solve for \(A\), \(B\), and \(C\).

Using the first three terms:
1. For \(n=1\): \(A(1^2) + B(1) + C = 4\) → \(A + B + C = 4\)
2. For \(n=2\): \(A(2^2) + B(2) + C = 14\) → \(4A + 2B + C = 14\)
3. For \(n=3\): \(A(3^2) + B(3) + C = 30\) → \(9A + 3B + C = 30\)

Now we have the system of equations:
1. \(A + B + C = 4\) (Equation 1)
2. \(4A + 2B + C = 14\) (Equation 2)
3. \(9A + 3B + C = 30\) (Equation 3)

### Step 3: Solve the system of equations
Subtract Equation 1 from Equation 2:
\[ (4A + 2B + C) - (A + B + C) = 14 - 4 \]
This simplifies to:
\[ 3A + B = 10 \quad (Equation 4) \]

Subtract Equation 2 from Equation 3:
\[ (9A + 3B + C) - (4A + 2B + C) = 30 - 14 \]
This simplifies to:
\[ 5A + B = 16 \quad (Equation 5) \]

Now, subtract Equation 4 from Equation 5:
\[ (5A + B) - (3A + B) = 16 - 10 \]
This simplifies to:
\[ 2A = 6 \quad \Rightarrow \quad A = 3 \]

Now substitute \(A = 3\) into Equation 4:
\[ 3(3) + B = 10 \quad \Rightarrow \quad 9 + B = 10 \quad \Rightarrow \quad B = 1 \]

Now substitute \(A = 3\) and \(B = 1\) into Equation 1:
\[ 3 + 1 + C = 4 \quad \Rightarrow \quad C = 0 \]

Thus, the nth term is:
\[ a_n = 3n^2 + n \]

### Step 4: Find the sum to n terms
The sum \(S_n\) of the first n terms is given by:
\[ S_n = \sum_{i=1}^{n} a_i = \sum_{i=1}^{n} (3i^2 + i) \]
This can be separated into two sums:
\[ S_n = 3\sum_{i=1}^{n} i^2 + \sum_{i=1}^{n} i \]

Using the formulas for the sums:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \]

Substituting these into the equation for \(S_n\):
\[ S_n = 3 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]
\[ = \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \]
\[ = \frac{n(n+1)(2n+1 + 1)}{2} \]
\[ = \frac{n(n+1)(2n+2)}{2} \]
\[ = n(n+1)(n+1) \]
\[ = n(n+1)^2 \]

### Final Answer
Thus, the sum to n terms of the series \(4 + 14 + 30 + 52 + 82 + 114 + \ldots\) is:
\[ S_n = n(n+1)^2 \]