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Question
Find `sum_("r" = 1)^"n"(3"r"^2 - 2"r" + 1)`
Solution
`sum_("r" = 1)^"n"(3"r"^2 - 2"r" + 1)`
= `3sum_("r" = 1)^"n" "r"^2 - 2 sum_("r" = 1)^"n""r" + sum_("r" = 1)^"n" 1`
= `3*("n"("n" + 1)(2"n" + 1))/6 - 2*("n"("n" + 1))/2 + "n"`
= `"n"[(("n" + 1)(2"n" + 1))/2 - ("n" + 1) + 1]`
= `"n"/2(2"n"^2 + 3"n" + 1 - 2"n" - 2 + 2)`
= `"n"/2(2"n"^2 + "n" + 1)`.
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