题目

设数列{an}、{bn}各项都是正数,a1=1,b1=2,若lgbn,lgan+1,lgbn+1成等差数列,5an,5bn,5an+1成等比数
列,求an,bn通项公式

提问时间：2020-10-09

答案

lgbn,lgan+1,lgbn+1成等差数列,5an,5bn,5an+1成等比数就得到①(an+1)^2=bn*bn+1②bn^2=an*an+1①*①(an+1)^4=bn^2*(bn+1)^2=an*(an+1)^2*(an+2)(an+1)^2=an*an+2所以an是个等比数列.由②可得b1^2=a1*a2那么a2=4∴an=...