Question

Let $T_n$ be the $n^{\text{th}}$ term and $S_n$ be the sum of $n$ terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\cdots n\text{terms}$. Then which of the following is/are true?

• A. $T_{10}=\frac{121}{4}$
• B. $S_{10}=\frac{505}{6}$
• C. $\sum _{r=1}^{10}\sqrt{T_r}=33$
• D. $S_{199}-S_{198}={10}^4$

Answer 1

Answer: (A), (D)

The correct options are
A $T_{10}=\frac{121}{4}$
D $S_{199}-S_{198}={10}^4$

Given series is $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\cdots n\text{terms}$

Now, the $n^{th}$ term of the series is
$T_n=\frac{1^3+2^3+\cdots +n^3}{1+3+5+\cdots +(2n-1)}\Rightarrow T_n=\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{n^2}(\because \text{Sum of odd natural numbers upto}2n-1\text{is}{n}^2)\Rightarrow T_n=\frac{1}{4}(n^2+2n+1)$

Now, $S_n=\sum _{r=1}^n{T}_r$
$\Rightarrow S_n=\frac{1}{4}\sum _{r=1}^n[n^2+2n+1]$
$\Rightarrow S_n=\frac{1}{4}[\frac{n(n+1)(2n+1)}{6}+n(n+1)+n]\Rightarrow S_n=\frac{n}{24}\left[\right(n+1\left)\right(2n+7)+6]\Rightarrow S_n=\frac{n}{24}[2n^2+9n+13]$

Therefore, $T_{10}=\frac{121}{4}$
$S_{10}=\frac{10}{24}(200+90+13)\Rightarrow S_{10}=\frac{5}{12}\times 303=\frac{505}{4}$

$\sum _{r=1}^{10}\sqrt{T_r}=\sum _{r=1}^{10}\frac{(r+1)}{2}\Rightarrow \sum _{r=1}^{10}\sqrt{T_r}=\frac{1}{2}[2+3+\cdots +11]\Rightarrow \sum _{r=1}^{10}\sqrt{T_r}=\frac{1}{2}[\frac{11\times 12}{2}-1]=\frac{65}{2}{S}_{199}-S_{198}=T_{199}\Rightarrow S_{199}-S_{198}=\frac{1}{4}(200{)}^2={10}^4$