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Question

1.25+2.3.6+3.4.7+....


Solution

1.25+2.3.6+3.4.7+....Let Tn be the nth term of this series.Then,Tn=(nth term of 1,2,3...)×(nth term of 2,3,4...)×(nth term of 5,6,7...)=[1+(n1)×1]×[2+(n1)×1]×[5+(n1)×1]=[1+n1]×[2+n1]×[5+n1]=n×(n+1)(n+4)=n3+4n2+n2+4n=n3+5n2+4n=Tn=n3+5n2+4nLet Snbe the sum to n terms of the given series; Then,Sn=nn1Tn=nn1(n3+5n2+4n)=nn1n3+nn15n2+nn14n=nn3+5nn1n=[n(n+1)2]2+5[n(n+1)(2n+1)2]+4[n(n+1)2]=n2(n+1)24+5n(n+1)(2n+1)6+2n(n+1)=3n2(n+1)2+10n(n+1)(2n+1)+24n(n+1)12=n(n+1)12[3n2+3n+20n+10+24]=n(n+1)12[3n2+23n+34]=n(n+1)(3n2+23n+34)12Hence, Sn =n(n+1)(3n2+23n+34)12


Mathematics
RD Sharma
Standard XI

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