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Question
Find the sum `sum_("r" = 1)^"n"("r" + 1)(2"r" - 1)`.
Solution
`sum_("r" = 1)^"n"("r" + 1)(2"r" - 1)`
= `sum_("r" = 1)^"n"(2"r"^2 + "r" - 1)`
= `2 sum_("r" = 1)^"n""r"^2 + sum_("r" = 1)^"n""r" - sum_("r" = 1)^"n"1`
= `2.("n"("n" + 1)(2"n" + 1))/6 + ("n"("n" + 1))/2 - "n"`
= `"n"/6[2(2"n"^2 + 3"n" + 1) + 3("n" + 1) - 6]`
= `"n"/6(4"n"^2 + 6"n" + 2 + 3"n" + 3 - 6)`
= `"n"/6 (4"n"^2 + 9"n" - 1)`.
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