1+(1+2)+(1+2+3)+(1+2+3+4)+...
Let Tn be the nth term of this series. Then, Tn=1+2+3+...+n=n2[2×1+(n−1)×1]=n12[2+n−1]=n12(n+1)=n22+n2
Let Sn denote the sum to n terms of the given series; Then,
∑nk−1Tk=∑nk−1[k22+k2]=∑nk−1k22+∑nk−1k2⇒Sn=12∑nk−1k2+12∑nk−1k=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