Question

The sum of all positive integers n for which $\frac{1^3+2^3+....+(2n{)}^3}{1^2+2^2+....+n^2}$ is also an integer is$?$

• A. $8$
• B. $9$
• C. $15$
• D. Infinite

Answer 1

Answer: (A)

$8$

We know that sum of cubes of natural numbers up to $k$ is given by $\frac{k^2(k+1{)}^2}{4}$ and

sum of squares of natural numbers up to $k$ is given by $\frac{k(k+1)(2k+1)}{6}$

Thus,

$\frac{1^3+2^3+...+(2n{)}^3}{1^2+2^2+...+n^2}=\frac{\left(2n{)}^2\right(2n+1{)}^2}{4}\times \frac{6}{n(n+1)(2n+1)}=\frac{6n(2n+1)}{n+1}$

We know that $n+1$ is not a factor of $2n+1$ and $n$ for $n\in N$

Thus, the given fraction is an integer, if $n+1$ is a factor of $6$.

This is possible for $n=1,2$ and $5$

Sum of all such $n$ is $1+2+5=8$