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Question

1×2+2×3+3×4+4×5+...


Solution

Let Tn be the nth term of this series. Then, Tn= (nth term of 1,2,3...)× (nth term of 2, 3, 4...) =[1+(n1)×1].[2+(n1)×1]=[1+n1].[2+n1]=n(n+1)=n2+nLet Sn be the sum to n terms of the given series; Then,Sn=nn1Tn=nn1(n2+n)=nn1n2+nn1n=n(n+1)(2n+1)6+(n+1)n2=n(n+1)(2n+1)+3n(n+1)6=n(n+1)[2n+1+3]6=n(n+1)[2n+4]6=n(n+1)×2(n+2)6=2n6(n+1)(n+2)Hence, Sn=n3(n+1)(n+2)


Mathematics
RD Sharma
Standard XI

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