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