Use the identities to evaluate the sums below: `sum_(i=1)^n(8i^3+2)/n^4*sum_(i=1)^n(12i^2-10)/n^3`

`
`

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

Get 48 Hours Free Access
Approved by eNotes Editorial