For which $\displaystyle n$ does $\displaystyle \sum_{i=1}^ni$ divide $\displaystyle \sum_{i=1}^n(i-1)i$ ?
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Take note that $\displaystyle \frac{\sum_{i=1}^{n}i}{\sum_{i=1}^{n}(i-1)i}=\frac{\frac{n(n+1)}{2}}{\frac{n(n-1)(n+1)}{3}}=\frac{3}{2(n-1)}$
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