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 $n = 1 + 2 + 5=8$