已知正项数列{an},{bn}满足:对任意正整数n,都有an,bn,a(n+1)成等差数列,bn,a(n+1),b（n＋1）成等比数列

题目

已知正项数列{an},{bn}满足:对任意正整数n,都有an,bn,a(n+1)成等差数列,bn,a(n+1),b（n＋1）成等比数列
且a1＝10,a2＝15,（1）求数列{√bn}是等差数列,（2）求{an}{bn}的通项公式（3）设Sn＝1/a1＋1/a2＋.＋1/an,如果对任意正整数n,不等式2aSn＜2－bn/an,求实数a的取值范围

答案

1、an,bn,a(n+1),所以,2bn=an+a(n+1)推出,2(bn+1)=a(n+1)+a(n+2)bn,a(n+1),b(n+1),所以,a(n+1)^2=bn*b(n+1),推出,a(n+1)=根号下【bn*b(n+1)】,a(n+2)=根号下【b(n+1)*b(n+2)】,带入2(bn+1)=a(n+1)+a(n+2)并化简,得到...