In the given series, there is sum of multiple of corresponding terms of two A.P.’s. The two A.P.’s are (i) 3, 6, 9, ............ and (ii) 8, 11, 14, ......... Now the nth term of sum is $a_n$ = (nth term of the sequence formed by first A.P.) × (nth term of the sequence formed by second A.P.) = (3n) (3n + 5) = $9n^2$ + 15n Hence, the sum to n terms is $S_n$ = $\sum\limits_{k=1}^{n} a_k$ = $\sum\limits_{k=1}^{n}(9k^2+15k)$ = $9\left[\frac{n(n+1)(2n+1)}{6}\right]$ + $15\left[\frac{n(n+1)}{2}\right]$ = $\frac{n(n+1)}{2}$ [3 (2n + 1) + 15] = $\frac{n(n+1)(6n+18)}{2}$ = 3n (n + 1) (n + 3)