Question

The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
[3 Marks]

Answer 1

$1^2+2^2+3^2+...+n^2=285$
$\Rightarrow \frac{n(n+1)(2n+1)}{6}=285$ ---(i) [0.5 Marks]

$1^3+2^3+3^3+...+n^3=2025$

$\Rightarrow {\left[\frac{n(n+1)}{2}\right]}^2=2025$ [0.5 Marks]

$\Rightarrow \frac{n(n+1)}{2}=\sqrt{2025}$

$\Rightarrow \frac{n(n+1)}{2}=45$

$\Rightarrow n(n+1)=90$ [1 Mark]

Substitute the value of $n(n+1)$ in (i),

$\Rightarrow \frac{90(2n+1)}{6}=285$

$\Rightarrow 2n+1=\frac{285\times 6}{90}$

$\Rightarrow 2n+1=19$
$\Rightarrow 2n=18$
$\Rightarrow n=9$
$\therefore$ The value of $n$ is 9. [1 Mark]