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)(2"n" + 1))/6 + 6("n"("n" + 1))/2`

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

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

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