Question

How do you use the properties of summation to evaluate the sum of `Sigma i(i-1)^2` from i=1 to 15?

Answer 1

`sum_(i=1)^15 i(i-1)^2 = 12040`

Explanation:

We need to use these the standard results:
`sum_(r=1)^n r = 1/2n(n+1)`
`sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1)`
`sum_(r=1)^n r^3 = 1/4n^2(n+1)^2`

We have;

`sum_(i=1)^n i(i-1)^2 = sum_(i=1)^n i(i^2-2i+1)`
`:. sum_(i=1)^n i(i-1)^2 = sum_(i=1)^n (i^3-2i^2+i)`
`:. sum_(i=1)^n i(i-1)^2 = sum_(i=1)^n i^3 - 2sum_(i=1)^n i^2 + sum_(i=1)^n i`

So for the sum requested, using the standard results;

`sum_(i=1)^15 i(i-1)^2 = 1/2(15)(15+1)-(2)(1/6)(15)(15+1)(30+1) + [1/2(15)(15+1)]^2`

`:. sum_(i=1)^15 i(i-1)^2 = 1/2(15)(16)-(1/3)(15)(16)(31) + [1/2(15)(16)]^2`
`:. sum_(i=1)^15 i(i-1)^2 = 120 - 2480 + 120^2`
`:. sum_(i=1)^15 i(i-1)^2 = 120 - 2480 + 14400`
`:. sum_(i=1)^15 i(i-1)^2 = 12040`

Note:
We could equally derive a single formula the sum of `n` terms;

`sum_(i=1)^n i(i-1)^2 = 1/2n(n+1) -2 1/6n(n+1)(2n+1) + 1/4n^2(n+1)^2`
`:. sum_(i=1)^n i(i-1)^2 = 1/2n(n+1) - 1/3n(n+1)(2n+1) + 1/4n^2(n+1)^2`
`:. sum_(i=1)^n i(i-1)^2 = 1/12n(n+1) {6 - 4(2n+1) +3n(n+1) }`

`:. sum_(i=1)^n i(i-1)^2 = 1/12n(n+1) (6 - 8n-4 -3n^2 + 3n )`
`:. sum_(i=1)^n i(i-1)^2 = 1/12n(n+1) (3n^2 - 5n + 2 )`

And with `n=15` we get:

`sum_(i=1)^15 i(i-1)^2 = 1/12(15)(16) {3(15)(15) - 5(15) + 2 }`
`:. sum_(i=1)^15 i(i-1)^2 = (20) (675-75+2)`
`:. sum_(i=1)^15 i(i-1)^2 = (20) (602)`
`:. sum_(i=1)^15 i(i-1)^2 = 12040`, as before