题目内容
已知数列{an}中,a1=2,anan+1=an+an+1,则
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由anan+1=an+an+1 有an+1an+2=an+1+an+2 两式左右两边分别相减得(an+1-1)(an-an+2)=0,
易知an+1≠1,∴an=an+1,即数列{an}是周期为2的数列,将a1=2代入anan+1=an+an+1可得a2=2,再由周期性可知an="2. " ∴
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易知an+1≠1,∴an=an+1,即数列{an}是周期为2的数列,将a1=2代入anan+1=an+an+1可得a2=2,再由周期性可知an="2. " ∴

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