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I am trying to simplify $$\frac{\sum_{i=1}^n i(i+1)}{n(n-1)}$$.

I am not sure how to simplify ${\sum_{i=1}^n i(i+1)}$ part.

How can I simplify it?

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5 Answers 5

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You may know the wellknown results $$ \sum_{i=1}^n i=\frac{n(n+1)}{2}$$ and $$ \sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6},$$ from which you can combine your sum.

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$i(i+1) = 2{{i+1} \choose 2}$, and $\sum_{i=1}^n {{i+1}\choose 2} = {{n+2}\choose 3}$, because ${{n+2}\choose 3} - {{n+1}\choose 3} = {{n+1}\choose 2}$ by Pascal's triangle.

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A simple solution comes from

$$(i+1)^3-i^3=3i^2+3i+1=3i(i+1)+1$$

Thus

$$\sum_{i=1}^n(i+1)^3-i^3=\left( 3 \sum_{i=1}^n i(i+1) \right)+n \,.$$

The left hand side sum is telescopic.

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Use $\sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^n i = \dfrac{n(n+1)}{2}$.

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Hints:

$$\begin{align*}\sum_{k=1}^n k^2&=\frac{n(n+1)(2n+1)}6\\ \sum_{k=1}^n k&=\frac{n(n+1)}2\end{align*}$$

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