数学…求通项公式在等差数列{An}中…a1=1前n项和Sn满足条件S2n/Sn=4n+2/n+1.n=1,2…求{an}

S2n/Sn=(4n+2)/(n+1)=n*(4n+2)/[n*(n+1)]=2n*(2n+1)/[n*(n+1)] S2n = 2n*(2n+1)*k,Sn=n*(n+1)*k S1=1*(1+1)*k=A1=1,k=1/2,即Sn=n*(n+1)/2 An=Sn-S(n-1)=n*(n+1)/2-n*(n-1)/2=n 即An＝n Sn=n+n(n-1)d/2 S2n=2n+(2n)(2n-1)d/2 S2n/Sn=[2n+(2n)(2n-1)d/2]/[n+n(n-1)d/2]=(4n+2)/(n+1) (4+2(2n-1)d)/(2+(n-1)d)=(4n+2)/(n+1) 4d=4,4-2d=2,d=1,2-d=1 d=1 an=n