Question:
Find the sum to $n$ terms of the series $3 \times 8+6 \times 11+9 \times 14+\ldots$
Solution:
The given series is $3 \times 8+6 \times 11+9 \times 14+\ldots$
$a_{n}=\left(n^{\text {th }}\right.$ term of $\left.3,6,9 \ldots\right) \times\left(n^{\text {th }}\right.$ term of $\left.8,11,14, \ldots\right)$
$=(3 n)(3 n+5)$
$=9 n^{2}+15 n$
$\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(9 k^{2}+15 k\right)$
$=9 \sum_{k=1}^{n} k^{2}+15 \sum_{k=1}^{n} k$
$=9 \times \frac{n(n+1)(2 n+1)}{6}+15 \times \frac{n(n+1)}{2}$
$=\frac{3 n(n+1)(2 n+1)}{2}+\frac{15 n(n+1)}{2}$
$=\frac{3 n(n+1)}{2}(2 n+1+5)$
$=\frac{3 n(n+1)}{2}(2 n+6)$
$=3 n(n+1)(n+3)$