Question

What is the value of the `sum_(i=1)^n(2i^2 + i - 8)`?

Answer 1

`sum_(i=1)^n(2i^2+i-8) = 2/3n^3+3/2n^2-43/6n`

Explanation:

We will use the formulas for the sum of the first `n` natural numbers and the sum of the squares of the first `n` natural numbers:

• `sum_(i=1)^ni = (n(n+1))/2`

• `sum_(i=1)^ni^2 = (n(n+1)(2n+1))/6`

Then we have

`sum_(i=1)^n(2i^2+i-8) = sum_(i=1)^n2i^2+sum_(i=1)^ni+sum_(i=1)^n(-8)`

`=2sum_(i=1)^ni^2+sum_(i=1)^ni-sum_(i=1)^n8`

`=2(n(n+1)(2n+1))/6 + (n(n+1))/2 - 8n`

(at this point we simply perform some basic algebra to simplify)

`=2/3n^3+n^2+1/3n+1/2n^2+1/2n-8n`

`=2/3n^3+3/2n^2-43/6n`