Question

Sum to n terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+.....$ is

• A. $\frac{n}{24}(n^2+9n+14)$
• B. $\frac{n}{24}(2n^2+7n+15)$
• C. $\frac{n}{24}(2n^2+9n+13)$
• D. $\frac{n}{24}(n^2+11n+12)$

Answer 1

Answer: (C)

$\frac{n}{24}(2n^2+9n+13)$

We know,

${\sum }_{i=1}^{n}k=1+2+3+\dots +n=\frac{n(n+1)}{2}$ (sum of first n natural numbers)

${\sum }_{i=1}^n{k}^2=1^2+2^2+3^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$ (sum of squares of the first n natural numbers)

${\sum }_{k=1}^n{k}^3=1^3+2^3+3^3+\dots +n^3=\frac{n^2(n+1{)}^2}{4}$ (sum of cubes of the first n natural numbers).

$1+3+.....+(2n-1)={\sum }_{k=1}^n(2k-1)=n^2$sum of the first odd n natural numbers).

Let $t_r$ denote the rth term of the series, then

$t_r=\frac{1^3+2^3+.....+r^3}{1+3+.....+(2r-1)}=\frac{\frac{1}{4}{r}^2(r+1{)}^2}{r^2}=\frac{1}{4}(r+1{)}^2$

$\Rightarrow \sum _{r=1}^n{t}_r=\frac{1}{4}\sum _{r=1}^n(r+1{)}^2=\frac{1}{4}[\sum _{r=1}^{n+1}{r}^2+2r+1]$

$=\frac{1}{4}[\frac{\left(n\right)(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}+n]=\frac{1}{24}[2n^3+n^2+2n^2+n+6n^2+6n+6n]$

$=\frac{n}{24}(2n^2+9n+13)$