Question

Let the series be $\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+\dots .$ Then

(where $T_n$ and $S_n$ denote the $n^{th}$ term and sum upto $n^{th}$ term respectively.)

• A. $T_n=\frac{2n+3}{3}$
• B. $S_n=\frac{n(n+2)}{3}$
• C. $S_{31}=420$
• D. $T_{31}=21$

Answer 1

Answer: (B), (D)

The correct options are
B $S_n=\frac{n(n+2)}{3}$
D $T_{31}=21$

$T_n=\frac{1^2+2^2+\cdots +n^2}{1+2+\cdots +n}$
$=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}$
$=\frac{n(n+1)(2n+1)}{3n(n+1)}$
$=\frac{1}{3}(2n+1)$
$\therefore T_{31}=21$

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

$S_n=\frac{n(n+2)}{3}$
$\therefore S_{31}=\frac{31\times 33}{3}=341$