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Question
Find the sum `sum_(r = 1)^n(r + 1)(2r - 1)`.
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Solution
`sum_(r = 1)^n(r + 1)(2r - 1)`
= `sum_(r = 1)^n(2r^2 + r - 1)`
= `2 sum_(r = 1)^nr^2 + sum_(r = 1)^nr - sum_(r = 1)^n1`
= `2.(n(n + 1)(2n + 1))/6 + (n(n + 1))/2 - "n"`
= `n/6[2(2n^2 + 3n + 1) + 3(n + 1) - 6]`
= `n/6(4n^2 + 6n + 2 + 3n + 3 - 6)`
= `n/6 (4n^2 + 9n - 1)`.
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