# Can someone give me a hint on how to prove this?

I'm supposed to prove that, for every integer $$n > 0,$$ it is true that $$(1 + 2 + ... + n)$$ divides $$3(1^2 + 2^2 + ... + n^2)$$.

Should I use induction?

This was given as an exercise in a chapter about divisibility in my algebra textbook.

$$1+2+...+n=\frac{n(n+1)}{2}$$
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
$$3(1^2+2^2+...+n^2)=3(\frac{1}{6}n(n+1)(2n+1))$$ $$=\frac{1}{2}n(n+1)(2n+1)$$ $$1+2+...+n=\frac{1}{2}n(n+1)$$ $$\therefore \frac{3(1^2+2^2+...+n^2)}{1+2+...+n}=\frac{\frac{1}{2}n(n+1)(2n+1)}{\frac{1}{2}n(n+1)}=2n+1$$ So, now we can see that $$3(1^2+2^2+...+n^2) =(2n+1)(1+2+...+n)$$ and hence $$(1+2+...+n)|3(1^2+2^2+...+n^2)$$