Sum the series $3\times 8 + 6\times 11 + 9\times 14 + \ldots$ to $n$ terms.

The two factors form arithmetic progressions: $3,6,9,\ldots$ and $8,11,14,\ldots$. So the nth term is $$t_n=(3n)(3n+5)=9n^2+15n.$$Therefore $$S_n=\sum_{k=1}^n (9k^2+15k)=9\sum_{k=1}^n k^2+15\sum_{k=1}^n k.$$Substituting the standard formulas, $$S_n=9\cdot \frac{n(n+1)(2n+1)}{6}+15\cdot \frac{n(n+1)}{2}.$$Simplifying gives $$S_n=\frac{3}{2}n(n+1)\big[(2n+1)+5\big]=3n(n+1)(n+3).$$