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Question

Find the sum of n terms of the series whose nth term is n3+32n.

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Solution

Tn=n3+32n

Let Sn denotes the sum of the n terms of the given series, then

Sn=nk=1Tk=nk=1(k3+32k)

=(nk=1k3)+32(nk=1k)

=[n(n+1)2]2+32[n(n+1)2]

=[n(n+1)2][n(n+1)2+32]

=n(n+1)(n2+n+3)4

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