Question

The sum of $n$ terms of the series $2\cdot 5+5\cdot 8+8\cdot 11+\dots$ is

• A. $3n(n-1{)}^2+n$
• B. $3n(n+1{)}^2-2n$
• C. $3n(n-1{)}^2+2n$
• D. $3n(n-1{)}^2-2n$

Answer 1

The correct option is B $3n(n+1{)}^2-2n$

Let
$S=2\cdot 5+5\cdot 8+8\cdot 11+\dots \dots$
So, the general term is,
$T_r=(3r-1)(3r+2)\Rightarrow T_r=9r^2+3r-2$

$S_n=\sum _{r=1}^n{T}_r=9\sum _{r=1}^n{r}^2+3\sum _{r=1}^{n}r-\sum _{r=1}^{n}2$
$=9\left(\frac{n(n+1)(2n+1)}{6}\right)+3\left(\frac{n(n+1)}{2}\right)-2n=\frac{3n(n+1)}{2}[2n+1+1]-2n$
$=3n(n+1{)}^2-2n$


Alternate method:
Putting $n=1$,
Then sum of $1$ term $=10$
$3n(n-1{)}^2+n=13n(n+1{)}^2-2n=103n(n-1{)}^2+2n=23n(n-1{)}^2-2n=-2$

Hence, the sum of $n$ terms is
$3n(n+1{)}^2-2n$