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Good proof of the formula $1+2+…+n=\frac{n(n+1)}{2}$ but the main task for the problem was to prove that $\frac{n(n+1)}{2}$ is always an integer even though there is a fraction over $2$. For $S_{n}=1^2+2^2+…+n^2$ consider the fact that $S_{n+1}-S_{n}=(n+1)^2$. If you can’t prove the sum of squares formula, then just show that $\frac{n(n+1)(2n+1)}{6}$ is always an integer for integer $n$.

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