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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.

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You can start from proving by induction the well known formulas:

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

$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

If you know that then the exercise is very easy.

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  • $\begingroup$ Thanks, i forgot the second formula i think that will do $\endgroup$ – ArielK Feb 9 at 21:58
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$$3(1^2+2^2+...+n^2)=3(\frac{1}{6}n(n+1)(2n+1))$$ $$=\frac{1}{2}n(n+1)(2n+1)$$ $$1+2+...+n=\frac{1}{2}n(n+1)$$ $$\therefore \frac{3(1^2+2^2+...+n^2)}{1+2+...+n}=\frac{\frac{1}{2}n(n+1)(2n+1)}{\frac{1}{2}n(n+1)}=2n+1$$ So, now we can see that $3(1^2+2^2+...+n^2) =(2n+1)(1+2+...+n)$ and hence $$(1+2+...+n)|3(1^2+2^2+...+n^2)$$

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