Question

Find \[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\].

Answer 1

\[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\]

= \[\displaystyle\sum_{r=1}^{n}(r^3 - 5r^2 + 6r)\].

= \[\displaystyle\sum_{r=1}^{n}r^3 - 5\displaystyle\sum_{r=1}^{n}r^2 + 6\displaystyle\sum_{r=1}^{n}r\]

= `(n^2(n + 1)^2)/4 - 5(n(n + 1)(2n + 1))/6 + 6(n(n + 1))/2`

= `(n(n + 1))/12[3n(n + 1) - 10(2n + 1) + 36]`

= `(n(n + 1))/12(3n^2 + 3n - 20n - 10 + 36)`

= `(n(n + 1))/12(3n^2 - 17n + 26)`