# Show that summation 1*2+2*3+...+(n-1)*n=(n-1)(n)(n+1)/3

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You need to prove that `sum_(k=2)^n` `(k-1)*k = ((n-1)(n)(n+1))/3`

Removing the brackets the summation will be: `sum_(k=2)^n ` `k^2` - `sum_(k=2)^n` `k` `sum_(k=2)^n` `k^2 =2^2 + 3^2 + ......+ n^2 = (n(n+1)(2n+1))/6 - 1`

`sum_(k=2)^n` k = `2 + 3 + .... + n = (n(n+1))/2 - 1`

`` `sum_(k=2)^n` (k-1)k `=(n(n+1)(2n+1))/6 - 1 - (n(n+1))/2+ 1`

Remove opposite members:

`(n(n+1)(2n+1))/6 - (n(n+1))/2 = [n(n+1)(2n+1)-3n(n+1)]/6`

Factoring n(n+1) => `(n(n+1)(2n+1-3))/6 = (n(n+1)(2n-2))/6`

`(n(n+1)(2n-2))/6 = (2n(n+1)(n-1))/6 = (n(n+1)(n-1))/3`

`` `sum_(k=2)^n` (k-1)k = `(n(n+1)(n-1))/3`

**The last line proves that `1*2 + 2*3 + .... + (n-1)*n = (n(n+1)(n-1))/3.` **