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Question

1.32+2.52+3.72+.... to 20 terms.

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Solution

General term Tr=r(3+(r1)2)2=r(2r+1)2=4r3+4r2+r
Sum =nr=1Tr=4nr=1r3+4nr=1r2+nr=1r
=4(n(n+1)2)2+4n(n+1)(2n+1)6+n(n+1)2
=16n(n+1)(6n2+14n+7)
By putting n=20
we get required sum =1620(20+1)(6×202+14×20+7)=188090

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