Evaluate `sum_(i=1)^n (8i^3+2)/n^4 * sum_(i=1)^n(12i^2-10)/n^3` :

First we recognize that `n` is a constant, and `sumk(f(i))=ksumf(i)`

So we can rewrite as :

`1/n^4sum_(i=1)^n(8i^3+2)*1/n^3sum_(i=1)^n(12i^2-10)`

Then we note that `sum(f+g)=sumf+sumg` .

`[1/n^4(sum_(i=1)^n8i^3+sum_(i=1)^n2)][1/n^3(sum_(i=1)^n12i^2-sum_(i=1)^n10)]`

Again we can pull out the constants. Also, `sum_(i=1)^na=an` where `a` is a constant. Thus we get:

`[1/n^4(8sum_(i=1)^ni^3+2n)][1/n^3(12sum_(i=1)^ni^2-10n)]`

Now we use the summation rules `sum_(i=1)^ni^2=n^3/3+n^2/2+n/6` and `sum_(i=1)^ni^3=n^4/4+n^3/2+n^2/4` :

`[1/n^4(8(n^4/4+n^3/2+n^2/4)+2n)][1/n^3(12(n^3/3+n^2/2+n/6)-10n)]`

`=[1/n^4(2n^4+4n^3+2n^2+2n)][1/n^3(4n^3+6n^2+2n-10n)]`

`=[2+4/n+2/n^2+2/n^3][4+6/n-8/n^2]`

** If you really meant to multiply here, you can use the distributive
property. **If you meant to find the two sums then:**

---------------------------------------------------------

`sum_(i=1)^n(8i^3+2)/n^4=2+4/n+2/n^2+2/n^3` and `sum_(i=1)^n(12i^2-10)/n^3=4+6/n-8/n^2`

--------------------------------------------------------

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.

**Further Reading**