Question

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

Answer 1

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

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

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

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

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

= `"n"/2(2"n"^2 + "n" + 1)`.