Question

已知公差不为0的等差数列an各项均为正数其前n项和为Sn满足2S2=a2(a2+1) ,a1、a2、a4成等比数列

Answer 1

(1)

∵a1、a2、a4成等比数列

∴a²2=a1*a4

即(a1+d)²=a1*(a1+3d)

a²1+2a1d+d²=a²1+3a1d

a1d=d²

∵d>0

∴a1=d a2=2d

2S2=2(a1+a2)=a2(a2+1)

2(d+2d)=2d(2d+1) (d>0)

解得 d=1

∴an=1+(n-1)x1=n

(2)

Sn=(1+n)n/2

bn=(2Sn+13)/n=n+13/n +1≥ 2√13 +1

最小值在n=13/n 即 n=√13的时取得,但是n为整数,考虑到 3