已知数列{an}是等差数列,a1=2,且a2,a3,a4+1成等比数列

已知数列{an}是等差数列,a1=2,且a2,a3,a4+1成等比数列

设公差为b,(2+b)/(2+2b)=(2+2b)/(3+3b),所以b=2,即an=2nbn=2/n*(2n+2)=1/n*(n+1)bn=(1/n)*1/(n+1)=1/n-1/(n+1)所以sn=b1+b2...+bn=1/n-1/(n+n)=1/2n