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Question

3×12+5×22+7×32+...


Solution

3×12+5×22+7×32+...Let Tn be the nth term of this series.Then,Tn=(nth term of 3,5,7...)×(nth term of 12,22,32...)=[3+(n1)2].[n2]=[3+2n2].(n2)=[2n+1][n2]2n3+n2Let Sn denote the sum of n terms of the given series; Then,Sn=nn1Tn=nn1(2n3+n2)=nn12n3+nn1n2=2nn1n3+nn1n2=2[n(n+1)2]2+[n(n+1)(2n+1)6]=24[n(n+1)]2+[n(n+1(2n+1))]6=[n(n+1)]22+n(n+1)(2n+1)6=n(n+1)6[3n(n+1)+(2n+1)]=n(n+1)6[3n2+3n+2n+1)]=n6(n+1)(3n2+5n+1)Hence, Sn = n6(n+1)(3n2+5n+1)


Mathematics
RD Sharma
Standard XI

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