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Question
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + ... + (2n – 1) (2n + 1) (2n + 3)
Solution
1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + ... + (2n – 1) (2n + 1) (2n + 3)
= `sum_("r" = 1)^"n" (2"r" - 1)(2"r" + 1)(2"r"+ 3)`
= `sum_("r" = 1)^"n" (4"r"^2 - 1)(2"r" + 3)`
= `sum_("r" = 1)^"n"(8"r"^3 + 12"r"^2 - 2"r" - 3)`
= `8sum_("r" = 1)^"n" "r"^3 + 12 sum_("r" = 1)^"n" "r"^2 - 2 sum_("r" = 1)^"n" "r" - 3 sum_("r" = 1)^"n" 1`
`=8.("n"^2("n" + 1)^2)/4 + 12*("n"("n" + 1)(2"n" + 1))/6 - 2.("n"("n" + 1))/2 - 3"n"`
= 2n2(n + 1)2 + 2n(n + 1)(2n + 1) – n(n + 1) – 3n
= n[2n(n2 + 2n + 1) + 2(2n2 + 3n + 1) – (n + 1) – 3]
= n(2n3 + 4n2 + 2n + 4n2 + 6n + 2 – n – 1 – 3)
= n(2n3 + 8n2 + 7n – 2)
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