Question

The sum of first $9$ terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+....$ is

• A. $71$
• B. $96$
• C. $142$
• D. $192$

Answer 1

The correct option is B $96$

$r^{th}$ term of the given series is, $T_r=\frac{1^3+2^3+3^3+.......+r^3}{1+3+5+.....+2r-1}$

[By using ${\sum }_{i=1}^n{k}^3=1^3+2^3+3^3+\dots +n^3=\frac{n^2(n+1{)}^2}{4}$ and $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ we have

$\Rightarrow T_r=\frac{{\left(\frac{r(r+1)}{2}\right)}^2}{\frac{r}{2}(2+(r-1\left)2\right)}$

$=\frac{1}{4}(r+1{)}^2$

$=\frac{1}{4}(r^2+2r+1)$

Thus sum of $n$ term is,

$\sum T_r=\frac{1}{4}[\sum r^2+2\sum r+\sum 1]$

$=\frac{1}{4}(\frac{n(n+1)(2n+1)}{6}+n(n+1)+n)$

$=\frac{1}{4}(\frac{n(n+1)(2n+1)}{6}+2\frac{1}{2}n(n+1)+n)$

Putting $n=9$ we get required sum

$=\frac{1}{4}(\frac{9\left(10\right)\left(19\right)}{6}+9(10)+9)$

$=96$