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Question
Find `sum_("r" = 1)^"n" (1^3 + 2^3 + ... + "r"^3)/("r"("r" + 1)`.
Solution
We know that,
13 + 23 + 33 + ... + n3 = `("n"^2("n" + 1)^2)/4`
∴ 13 + 23 + 33 + ... + r3 = `("r"^2("r" + 1)^2)/4`
∴ `(1^3 + 2^3 + 3^3 + .... + "r"^3)/("r"("r" + 1)) = ("r"("r" + 1))/4`
∴ `sum_("r" = 1)^"n"[(1^3 + 2^3 + 3^3 + .... + "r"^3)/("r"("r" + 1))]`
= `sum_("r" = 1)^"n"("r"("r" + 1))/4`
= `1/4 sum_("r" = 1)^"n"("r"^2 + "r")`
= `1/4(sum_("r" = 1)^"n" "r"^2 + sum_("r" = 1)^"n""r")`
= `1/4[("n"("n" + 1)(2"n" + 1))/6 + ("n"("n" + 1))/2]`
= `1/4.("n"("n" + 1))/2 ((2"n" + 1)/3 + 1)`
= `("n"("n" + 1))/8 ((2"n" + 1 + 3)/3)`
= `("n"("n" + 1)(2"n" + 4))/24`
= `(2"n"("n" + 1)("n" + 2))/24`
= `("n"("n" + 1)("n" + 2))/12`.
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