设数列{an}满足:a1=2,a2=5/3 ,an+2={(5/3)an+1}-(2/3)an ,n=1,2,3,…… 令bn=an+1-an (n=1,2,3,4, ……），求数列｛bn}的通项公式！
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设数列{an}满足:a1=2,a2=5/3 ,an+2={(5/3)an+1}-(2/3)an ,n=1,2,3,…… 令bn=an+1-an (n=1,2,3,4, ……）,求数列｛bn}的通项公式！ 解:an+2={(5/3)an+1}-(2/3)an .∴an+2={(an+1)+(2/3)an+1}-(2/3)an .∴an+2-(an+1)=[(2/3)an+1}-(2/3)an .an+2-(an+1)/(an+1)-an =2/3∵bn=an+1-an ∴bn+1/bn=2/3｛bn}是等比数列,b1=a2-a1=5/3-1=2/3,公比为2/3bn=(2/3)^n