Find the sum of 'n' terms of the series whose `n^(th)` term is `t_(n) = 3n^(2) + 2n`.

Answer

To find the sum of the first 'n' terms of the series whose nth term is given by \( t_n = 3n^2 + 2n \), we can follow these steps:

### Step 1: Write the expression for the sum of the first 'n' terms
The sum of the first 'n' terms, denoted as \( S_n \), can be expressed as:
\[ S_n = t_1 + t_2 + t_3 + \ldots + t_n = \sum_{k=1}^{n} t_k \]
Substituting the expression for \( t_k \):
\[ S_n = \sum_{k=1}^{n} (3k^2 + 2k) \]

### Step 2: Separate the summation
We can separate the summation into two parts:
\[ S_n = \sum_{k=1}^{n} 3k^2 + \sum_{k=1}^{n} 2k \]
This can be simplified to:
\[ S_n = 3\sum_{k=1}^{n} k^2 + 2\sum_{k=1}^{n} k \]

### Step 3: Use the formulas for the summation of squares and natural numbers
We will use the following formulas:
- The sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 4: Substitute the formulas into the expression for \( S_n \)
Substituting these formulas into our expression for \( S_n \):
\[ S_n = 3 \cdot \frac{n(n + 1)(2n + 1)}{6} + 2 \cdot \frac{n(n + 1)}{2} \]

### Step 5: Simplify the expression
Now we simplify each term:
1. The first term:
\[ S_n = \frac{3n(n + 1)(2n + 1)}{6} = \frac{n(n + 1)(2n + 1)}{2} \]
2. The second term:
\[ 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]

Combining these two results:
\[ S_n = \frac{n(n + 1)(2n + 1)}{2} + n(n + 1) \]

### Step 6: Factor out common terms
Factoring out \( n(n + 1) \):
\[ S_n = n(n + 1) \left( \frac{2n + 1}{2} + 1 \right) \]
This simplifies to:
\[ S_n = n(n + 1) \left( \frac{2n + 1 + 2}{2} \right) = n(n + 1) \left( \frac{2n + 3}{2} \right) \]

### Final Result
Thus, the sum of the first 'n' terms of the series is:
\[ S_n = \frac{n(n + 1)(2n + 3)}{2} \]