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Question

1+(1+2)+(1+2+3)+(1+2+3+4)+...


Solution

Let Tn be the nth term of this series. Then, Tn=1+2+3+...+n=n2[2×1+(n1)×1]=n12[2+n1]=n12(n+1)=n22+n2

Let Sn denote the sum to n terms of the given series; Then,

nk1Tk=nk1[k22+k2]=nk1k22+nk1k2Sn=12nk1k2+12nk1k=12[n(n+1)(2n+1)3]+12[n(n+1)2]=n(n+1)(2n+1)+3n(n+1)12 =n(n+1)12[2n+1+3]=n(n+1)12[2n+4]=n(n+1)12×2(n+2)=n6(n+1)(n+2)Hence,Sn=n6(n+1)(n+2)=n(n+1)2[2n+46]=n(n+1)×2[n+2]2×6


Mathematics
RD Sharma
Standard XI

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