Find the sum to n terms of the series whose nth term is
`3n^(2)+2n`

Answer

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

### Step 1: Write the expression for the sum of the first n terms
The sum \( S_n \) of the first n terms can be expressed as:
\[ S_n = \sum_{k=1}^{n} d_k = \sum_{k=1}^{n} (3k^2 + 2k) \]

### Step 2: Split 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 rewritten as:
\[ S_n = 3 \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k \]

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

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

### Step 5: Simplify the expression
Now we simplify each term:
\[ S_n = \frac{3n(n+1)(2n+1)}{6} + \frac{2n(n+1)}{2} \]
\[ S_n = \frac{3n(n+1)(2n+1)}{6} + n(n+1) \]
To combine these, we need a common denominator:
\[ S_n = \frac{3n(n+1)(2n+1)}{6} + \frac{6n(n+1)}{6} \]
\[ S_n = \frac{3n(n+1)(2n+1) + 6n(n+1)}{6} \]

### Step 6: Factor out common terms
Factoring out \( n(n+1) \):
\[ S_n = \frac{n(n+1)(3(2n+1) + 6)}{6} \]
\[ S_n = \frac{n(n+1)(6n + 9)}{6} \]

### Step 7: Final simplification
We can simplify this further:
\[ S_n = \frac{n(n+1)(3n + 4.5)}{3} \]
This gives us the final expression for the sum of the first n terms.

### Final Answer
Thus, the sum to n terms of the series is:
\[ S_n = \frac{n(n+1)(3n + 4.5)}{3} \]