Question

$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots .n$ terms $=$

• A. $\frac{n(2n^2+9n+13)}{24}$
• B. $\frac{n(2n^3+9n+13)}{8}$
• C. $\frac{n(n^2+9n+13)}{24}$
• D. $\frac{n(n^2+9n+13)}{8}$

Answer 1

The correct option is A $\frac{n(2n^2+9n+13)}{24}$

General $k^{th}$ term $=\frac{{\sum }_{i=1}^k{i}^3}{{\sum }_{i=1}^k(2i-1)}$ $=\frac{(k+1{)}^2}{4}$

Final answer : $\sum _{k=1}^n\frac{(k^2+2k+1)}{4}$

$⟹S_n=\frac{1}{4}\sum _{k=1}^n{k}^2+2k+1$

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)

$\therefore S_n=\frac{n(n+1)(2n+1)}{24}+\frac{n(n+1)}{4}+\frac{n}{4}$

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

Hence, option 'A' is correct.