an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)

显然在没有更多提示的情况下,等差数列的前n项和S(n)=a(1)*+d*n(n-1)/2=(d/2)*n^2+[a(1)-d/2]*n 因此S(n+1)=(d/2)*(n+1)^2+[a(1)-d/2]*(n+1)=(d/2)*n^2+[a(1)+d/2]*n+a(1) S(n-1)=(d/2)*(n-1)^2+[a(1)-d/2]*(n-1)=(d/2)*n^2+[a(1)-3d/2]*n-a(1)+d S(n)+S(n+1)=d*n^2+2a(1)*n+a(1) S(n)+S(n-1)=d*n^2+2[a(1)-d]*n-a(1)+d 作商取极限可得:lim[S(n)+S(n+1)]/[S(n)+S(n-1)] =[1+2a(1)/(dn)+a(1)/(dn^2)]/[1+2[a(1)-d]/(dn)+[-a(1)+d]/(dn^2)]=1