Question

How do you use the properties of summation to evaluate the sum of `Sigma (i^2-1)` from i=1 to 10?

Answer 1

`sum_(i=1)^10(i^2-1) = 375`

Explanation:

We could just add up the 10 terms but the numbers get a bit horrific, so it is actually easier in this case to derive a general formula standard formula for `sumi^2` we have:

`sum_(i=1)^n(i^2-1) = sum_(i=1)^n i^2 - sum_(i=1)^n 1`
`:. sum_(i=1)^n(i^2-1) = 1/6n(n+1)(2n+1) - n`
`:. sum_(i=1)^n(i^2-1) = 1/6n{(n+1)(2n+1) - 6}`
`:. sum_(i=1)^n(i^2-1) = 1/6n{2n^2+3n+1 - 6}`
`:. sum_(i=1)^n(i^2-1) = 1/6n{2n^2+3n- 5}`
`:. sum_(i=1)^n(i^2-1) = 1/6n(n-1)(2n+5)`

So with n=10 we have:
`sum_(i=1)^10(i^2-1) = 1/6(10)(9)(25)`
`sum_(i=1)^10(i^2-1) = 375`