+...+[(\frac{2n}{n} )^3-\frac{2n}{n} ](\frac{2}{n} )](/tpl/images/1854/7073/c667c.png)
Answers: 2
![\displaystyle \left[\left(\frac{2}{n}\right)}^3-\frac{2}{n}\right]\left(\frac{2}{n}\right)+...+\left[ \left(\frac{2n}{n}\right)^3-\frac{2n}{n}\right]\left(\frac{2}{n}\right)](/tpl/images/0830/1243/06074.png)
![\displaystyle =\frac{2}{n}\left(\left[\left(\frac{2}{n}\right)^3-\frac{2}{n}\right]+...+\left[\left(\frac{2n}{n}\right)^3-\frac{2n}{n}\right]\right)](/tpl/images/0830/1243/f3187.png)
. The last term is given by  
. Because at the last term, our initial term i is equal to n, we can replace the extra n with i for our summation.![=\displaystyle \sum_{i=1}^{n} \frac{2}{n}\left(\left[\left(\frac{2i}{n}\right)^3-\frac{2i}{n}\right]\right)](/tpl/images/0830/1243/29f56.png)
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