Question

The sum of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots \dots$ upto $16$ terms is

• A. $246$
• B. $646$
• C. $446$
• D. $746$

Answer 1

The correct option is C $446$

$T_n=\frac{\sum n^3}{\sum (2n-1)}$
$=\frac{1}{4}\times \frac{n^2(n+1{)}^2}{n^2}=\frac{1}{4}(n+1{)}^2\cdots (1)$
Now,
$S_{16}=\frac{1}{4}(2^2+3^2+4^2+\cdots {17}^2+1^2-1^2)=\frac{1}{4}(\frac{17(17+1)(2\cdot 17+1)}{6}-1)\left(\text{sum of squares of first}n\text{natural numbers}\right)=\frac{1}{4}(1785-1)=446$