Find the sigma notation of the following sum: [((2)/(n))^3-(2)/(n) ]((2)/(n) )+...+[((2n)/(n) )^3-(2n)/(n) ]((2)/(n) )

Question

Find the sigma notation of the following sum:


`[((2)/(n))^3-(2)/(n) ]((2)/(n) )+...+[((2n)/(n) )^3-(2n)/(n) ]((2)/(n) )`

Answer 1

`\displaystyle (2)/(n) \sum_(i = 1)^(n) \left(\left((2i)/(n)\right)^3 - (2i)/(n)\right)`

Explanation:

We have the sum:

`\displaystyle \left[\left((2)/(n)\right)}^3-(2)/(n)\right]\left((2)/(n)\right)+...+\left[ \left((2n)/(n)\right)^3-(2n)/(n)\right]\left((2)/(n)\right)`

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:

`\displaystyle =(2)/(n)\left(\left[\left((2)/(n)\right)^3-(2)/(n)\right]+...+\left[\left((2n)/(n)\right)^3-(2n)/(n)\right]\right)`

Notice the differences between the first term and the last term. The first term is given by
`\displaystyle \left((2)/(n)\right)^3- (2)/(n)`. The last term is given by
`\displaystyle \left((2n)/(n)\right)^3- (2n)/(n)`. 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:

`=\displaystyle \sum_(i=1)^(n) (2)/(n)\left(\left[\left((2i)/(n)\right)^3-(2i)/(n)\right]\right)`

And since (2 / n) is constant:

`\displaystyle = (2)/(n) \sum_(i = 1)^(n) \left(\left((2i)/(n)\right)^3 - (2i)/(n)\right)`