Answer:
[tex]\displaystyle \frac{2}{n} \sum_{i = 1}^{n} \left(\left(\frac{2i}{n}\right)^3 - \frac{2i}{n}\right)[/tex]
Step-by-step explanation:
We have the sum:
[tex]\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)[/tex]
And we want to find the sigma notation that represents the sum.
Note that the leftmost term is the first term while the rightmost term is the last term.
n is the number of terms. Therefore, (2 / n) will stay constant. Hence, we can factor it out of the series:
[tex]\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)[/tex]
Notice the differences between the first term and the last term. The first term is given by [tex]\displaystyle \left(\frac{2}{n}\right)^3- \frac{2}{n}[/tex]. The last term is given by [tex]\displaystyle \left(\frac{2n}{n}\right)^3- \frac{2n}{n}[/tex]. Because at the last term, our initial term i is equal to n, we can replace the extra n with i for our summation.
Therefore, we can write the following summation:
[tex]=\displaystyle \sum_{i=1}^{n} \frac{2}{n}\left(\left[\left(\frac{2i}{n}\right)^3-\frac{2i}{n}\right]\right)[/tex]
And since (2 / n) is constant:
[tex]\displaystyle = \frac{2}{n} \sum_{i = 1}^{n} \left(\left(\frac{2i}{n}\right)^3 - \frac{2i}{n}\right)[/tex]
