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Question
Find `sum_("r" = 1)^"n" [(1^3 + 2^3 + .... + "r"^3)/("r"("r" + 1))]`
Solution
`sum_("r" = 1)^"n" [(1^3 + 2^3 + .... + "r"^3)/("r"("r" + 1))]`
= `sum_("r"=1)^"n"[(sum_("i"=1)^"r""i"^3)/("r"("r"+1))]`
= `sum_("r" = 1)^"n" ("r"^2("r" + 1)^2)/(4"r"("r" + 1))`
= `sum_("r" = 1)^"n" ("r"("r" + 1))/4`
= `sum_("r" = 1)^"n" ("r"^2 + "r")/4`
= `sum_("r" = 1)^"n" (1/4"r"^2 + 1/4"r")`
= `1/4 sum_("r" = 1)^"n" "r"^2 + 1/4 sum_("r" = 1)^"n""r"`
= `1/4.("n"("n" + 1)(2"n" + 1))/6 + 1/4.("n"("n" + 1))/2`
= `("n"("n" + 1))/8 [(2"n" + 1)/3 + 1]`
= `("n"("n" + 1))/8[(2"n" + 1 + 3)/3]`
= `("n"("n" + 1)(2"n" + 4))/24`
= `("n"("n" + 1)("n" + 2))/12`
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