Question

Find the sum to n terms

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

Answer 1

The given series is

$3\times 3+6\times 11+9\times 14+\dots \dots$ to n terms.

Let '$a_n$' be the nth term of the given series and '$S_n$' be the sum of n terms of the given series.

$\therefore a_n=$ nth term of (3, 6, 9....) (nth term of 8, 11, 14 ....)

$=[3+(n-1)\times 3][8+(n-1)\times 3]$

$=3n(3n+5)=9n^2+15n$

$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(9k^2+15k)$

$=({9.1}^2+15.1)+({9.2}^2+15.2)+({9.3}^2+15.3)+\dots \dots +(9n^2+15n)$

$=9\left[\right(1^2+2^2+3^2+\dots \dots +n^2)+15(1+2+3+\dots \dots +n\left)\right]$

$=9.\frac{n(n+1)(2n+1)}{6}+\frac{15n(n+1)}{2}$

$=\frac{n(n+1)}{2}[6n+3+15]$

$=\frac{n(n+1)(6n+18)}{2}$

$=3n(n+1)(n+3)$