Correct Answer - Option 3 :

\(\rm \frac{n(3n+7)}{2}\)
__Concept:__

- Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
- Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
- Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

__Calculation:__

Given: nth term of a series is T_{n} = 3n + 2

Here, we have to find the value of \({S_n} = \;\mathop \sum \limits_{k = 1}^n {T_k} = \;?\)

Sum of series = \(\rm S_n = \sum (3n+2)=\sum 3n+\sum 2\)

As we know that, \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)

\(\Rightarrow \rm S_n =3\sum n+\sum 2\)

\(\Rightarrow \rm S_n =\frac{3n(n+1)}{2}+2n =\frac{n(3n+7)}{2}\)

Hence, **option 3** is the correct answer