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 integers is

• A. 8
• B. 9
• C. 15
• D. Infinite

Answer 1

Answer: (A)

Given,
$\frac{1^3+2^3+3^3+...+{\left(2n\right)}^3}{1^2+2^2+3^2+...+n^2}$
$=\frac{\frac{{\left(2n\right)}^2{\left(2n+1\right)}^2}{4}}{\frac{n\left(n+1\right)\left(2n+1\right)}{6}}$
$\left[\because \Sigma n^3=\frac{n^2{\left(n+1\right)}^2}{4},\Sigma n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]$
$=\frac{\frac{4n^2{\left(2n+1\right)}^2}{4}}{\frac{n\left(n+1\right)\left(2n+1\right)}{6}}=\frac{6n\left(2n+1\right)}{n+1}$
$\frac{12n^2+6n}{n+1}=\left(12n-6\right)+\frac{6}{n+1}$
$\because \frac{6}{n+1}$ is an integer if $n+1$ is factor of $6$.
$\because n+1=1,2,3,6$
$\Rightarrow =1,2,5$
Sum of $n=1+2+5=8$