等差数列{an}a1=1前n项和为Sn且S2n/Sn=4n+2/n+1 (1)求an通项公试

设公差为d
根据Sn=n(a1+an)/2＝na1+n(n-1)d/2＝n[2+(n-1)d]/2
S2n/Sn＝2[2+(2n-1)d]/[2+(n-1)d]
与原式比较,即：2[2+(2n-1)d]/[2+(n-1)d]=(4n+2)/(n+1 )
d=1
an=1+n-1＝n