Find the sum of first n terms of the series whose nth term is `3n^(2)+5`.

Answer

To find the sum of the first n terms of the series whose nth term is given by \( a_n = 3n^2 + 5 \), 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:
\[ S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (3k^2 + 5) \]

### Step 2: Split the sum into two parts
We can separate the sum into two distinct sums:
\[ S_n = \sum_{k=1}^{n} 3k^2 + \sum_{k=1}^{n} 5 \]
This simplifies to:
\[ S_n = 3 \sum_{k=1}^{n} k^2 + 5 \sum_{k=1}^{n} 1 \]

### Step 3: Use the formula for the sum of squares
The formula for the sum of the first n squares is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Substituting this into our equation gives:
\[ S_n = 3 \left(\frac{n(n + 1)(2n + 1)}{6}\right) + 5n \]

### Step 4: Simplify the expression
Now, we can simplify the first term:
\[ S_n = \frac{3n(n + 1)(2n + 1)}{6} + 5n = \frac{n(n + 1)(2n + 1)}{2} + 5n \]

### Step 5: Combine the terms
To combine the terms, we can express \( 5n \) with a common denominator:
\[ 5n = \frac{10n}{2} \]
Thus, we have:
\[ S_n = \frac{n(n + 1)(2n + 1) + 10n}{2} \]

### Step 6: Factor the numerator
Now, factor the numerator:
\[ S_n = \frac{n(n + 1)(2n + 1) + 10n}{2} = \frac{n[(n + 1)(2n + 1) + 10]}{2} \]

### Step 7: Final expression
The final expression for the sum of the first n terms is:
\[ S_n = \frac{n(n + 1)(2n + 1) + 10n}{2} \]

### Final Answer
Thus, the sum of the first n terms of the series is:
\[ S_n = \frac{n(n + 1)(2n + 1) + 10n}{2} \]

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