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
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