Question

$3\times 8+6\times 11+9\times 14+...$

Answer 1

$\text{The given series is},3\times 8+6\times 11+9\times 14+...a_n=(n^{th}\text{term of}3,6,\mathrm{9...})\times (n^{th}\text{term of}8,11,\mathrm{14...})=\left(3n\right)(3n+5)={9n}^2+15n\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n({9k}^2+15k)=9{\sum }_{k=1}^n{k}^2+15{\sum }_{k=1}^{n}k=9\times \frac{n(n+1)(2n+1)}{6}+15\times \frac{n(n+1)(2n+1)}{2}=\frac{3n(n+1)(2n+1)}{2}+\frac{15n(n+1)}{2}=\frac{3n(n+1)}{2}(2n+15)=\frac{3n(n+1)}{2}(2n+6)=3n(n+1)(n+3)$