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

Answer

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

### Step 1: Identify the nth term
The nth term of the series is given as:
\[ T_n = 3n + 2 \]

### Step 2: Write the sum of the first n terms
The sum of the first n terms \( S_n \) can be expressed as:
\[ S_n = T_1 + T_2 + T_3 + ... + T_n \]
This can also be written using sigma notation:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (3k + 2) \]

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

### Step 4: Calculate the sums
Using the formula for the sum of the first n natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
And the sum of 1 over n terms:
\[ \sum_{k=1}^{n} 1 = n \]
We can substitute these into our equation for \( S_n \):
\[ S_n = 3 \cdot \frac{n(n + 1)}{2} + 2n \]

### Step 5: Simplify the expression
Now, simplify the expression:
\[ S_n = \frac{3n(n + 1)}{2} + 2n \]
To combine the terms, we can express \( 2n \) with a common denominator:
\[ 2n = \frac{4n}{2} \]
Thus, we have:
\[ S_n = \frac{3n(n + 1) + 4n}{2} \]
Expanding the numerator:
\[ S_n = \frac{3n^2 + 3n + 4n}{2} = \frac{3n^2 + 7n}{2} \]

### Final Result
The sum of the series to n terms is:
\[ S_n = \frac{3n^2 + 7n}{2} \]