Question

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

Answer 1

`sum_(i=1)^10i(i^2+1) = 3080`

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` and `sumi^3` we have:

`sum_(i=1)^ni(i^2+1) = sum_(i=1)^n (i^3+i)`
`:. sum_(i=1)^ni(i^2+1) = sum_(i=1)^n i^3 + sum_(i=1)^n i`
`:. sum_(i=1)^ni(i^2+1) = (n^2(n+1)^2)/4 + (n(n+1))/2`
`:. sum_(i=1)^ni(i^2+1) = (n(n+1))/4 { n(n+1) + 2 }`
`:. sum_(i=1)^ni(i^2+1) = (n(n+1))/4 ( n^2 + n + 2)`

So with n=10 we have:
`:. sum_(i=1)^10i(i^2+1) = (10(11))/4 ( 100 + 10 + 2)`
`:. sum_(i=1)^10i(i^2+1) = (10(11)(112))/4`
`:. sum_(i=1)^10i(i^2+1) = 3080`