Question

30min内解决分数双倍设sn是等比数列{an}前n项,且s3,s9,s6成等差数列（1）若a2,a8,am成等差数列,

Answer 1

1.易得q≠1,a1≠0,2s9=s3+s6,解得q=(-2)^-1/3

2a8=a2+am,解得m=5

2.由1得bn=-n,cn=1/n*(n+1),裂项得Tn=1-1/n+1

假设存在m.n,有：Tm^2=Tn.T1

化简得m^2-2mn-n=0

(2n+2)^2＞4（n^2+n)＞（2n)^2,故无整数m.