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Question
Answer the following:
Find `sum_("r" = 1)^"n" ((1^2 + 2^2 + 3^2 + ... + "r"^2)/(2"r" + 1))`
Solution
`sum_("r" = 1)^"n" ((1^2 + 2^2 + 3^2 + ... + "r"^2)/(2"r" + 1))`
= `sum_("r" = 1)^"n" ((sum_("i" = 1)^"r" "i"^2)/(2"r" + 1))`
= `sum_("r" = 1)^"n" ("r"("r" + 1)(2"r" + 1))/(6(2"r" + 1))`
= `sum_("r" = 1)^"n" ("r"("r" + 1))/6`
= `sum_("r" = 1)^"n" ("r"^2 + "r")/6`
= `sum_("r" = 1)^"n" (1/6"r"^2 + 1/6"r")`
= `1/6 sum_("r" = 1)^"n" "r"^2 + 1/6 sum_("r" = 1)^"n" "r"`
= `1/6*("n"("n" + 1)(2"n" + 1))/6 + 1/6*("n"("n" + 1))/2`
= `("n"("n" + 1))/12 [(2"n" + 1)/3 + 1]`
= `("n"("n" + 1))/12 [(2"n" + 1 + 3)/3]`
= `("n"("n" + 1)(2"n" + 4))/36`
= `("n"("n" + 1)("n" + 2))/18`
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