Question

已知公差不为0的数列{an}的首相为a1=1，前n项的和为Sn，若数列{Sn/an}是等差数列

(1)求an
令bn=q的Sn次方（q＞0），若对一切n∈N*，都有bn+1^2＞2bn*bn+2,求q的取值范围

Answer 1

S1/a1=1
S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)
S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)

2*d/(1+d)=(2+d)/(1+2d)
解得d=1，d=0(舍去)
所以，an=n

(b(n+1))^2=q^(n+1)(n+2)=q^(n*n+3n+2)
2*bn*b(n+2)=2*q^{n(n+1)/2+(n+1)(n+2)/2}=2*q^(n*n+3n+3)

Answer 2

S1/a1=1
S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)
S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)

2*d/(1+d)=(2+d)/(1+2d)
解得d=1，d=0(舍去)
所以，an=n

(b(n+1))^2=q^(n+1)(n+2)=q^(n*n+3n+2)
2*bn*b(n+2)=2*q^{n(n+1)/2+(n+1)(n+2)/2}=2*q^(n*n+3n+3)