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Question

Sum the following series to n terms 1.4.7+4.7.10+7.10.13+.....

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Solution

General term=(3n2)(3n+1)(3n+4)
=27n3+27n218n8
Sum to nth term =27n3+27n2+18n8
=27n2(n+1)24+27(n)(2n+1)(n+1)618(n)(n+1)28
=n(n+1)[27(n2+n)4+27(2n+1)617]

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