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Question

Find the sum of the series 1.32+2.52+3.72+ to n terms.

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Solution

S=1.32+2.52+3.72+......S=nr=1r(2r+1)2=r(4r2+1+4r)=4r3+r+4r2=4[n(n+1)]24+n(n+1)2+4n(n+1)(2n+1)6=n2(n+1)2+n(n+1)2+23(n)(n+1)(2n+1)=n(n+1)[n(n+1)+12+23(2n+1)]=n(n+1)[n2+7n3+76]

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