Question

已知正项数列{an}中,a1=3,前n项和为Sn(n是N*)当n≥2时,有√Sn-√S(n-1)

Answer 1

1.

n≥2时S(n-1)+1anSn+1成等差数列则

2an=S(n-1)+1+Sn+1

2[Sn-S(n-1)]=S(n-1)+Sn+2

Sn=3S(n-1)+2

Sn+1=3S(n-1)+3=3[S(n-1)+1]

(Sn+1)/[S(n-1)+1]=3定值

S1+1=2+1=3

数列{Sn+1}3首项3公比等比数列

2...