令bn=1+2+3+...+n=n(n+1)/2=1/2[n^2+n],则Sn=b1+b2+...+bn=1/2[(1^2+1)+(2^2+2)+...+(n^2+n)]=1/2[(1^2+2^2+...+n^2)+(1+2+...+n)]=1/2[n(n+1)(2n+1)/6+n(n+1)/2]= n(n+1)(n+2)/6.其中1^2+2^2+...+n^2=n(n+1)(2n+1)/...