Question

The sum of the series $6+13+22+33+\dots \dots$ upto $20$ terms is

• A. $3730$
• B. $3720$
• C. $3710$
• D. $3750$

Answer 1

The correct option is A $3730$

Let
$S_n=6+13+22+33+\dots \dots +T_n$
$S_n=6+13+22+\dots \dots +T_{n-1}+T_n$
$\overline{0=6+7+9+11+\dots \dots +(T_n-T_{n-1})-T_n}$
or, $T_n=6+(7+9+11+\dots \dots +T_n-T_{n-1})$
$=6+\frac{(n-1)}{2}[2\times 7+(n-2\left)2\right]$
$=6+(n-1)(n+5)$
$\therefore T_n=n^2+4n+1$

Now, $S_n=\sum _{r=1}^n{T}_r=\sum _{r=1}^n{r}^2+4\sum _{r=1}^{n}r+\sum _{r=1}^{n}1$
$=\frac{n(n+1)(2n+1)}{6}+2n(n+1)+n$
$=\frac{n(n+1)}{6}[2n+1+12]+n=\frac{n(n+1)(2n+13)}{6}+n$
$\therefore S_{20}=3730$