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Question

Find the sum of n terms of the series whose nth term is n2(n21)4n21.

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Solution

Tn=n2(n21)4n21=n244n244n21=n2434n24n21=n243164n24n21
=n24316316(4n21)
=n24316316(2n1)(2n+1)
Therefore, Sn=Tn=n(n+1)(2n+1)243n163n16(2n+1)
=n(n+1)(2n+1)243n(n+1)8(2n+1
=(n1)n(n+1)(n+2)6(2n+1)

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