设数列a1,a2,…,an,…满足a1=a2=1,a3=2,且对任何自然数n,都有anan+1an+2≠1,又anan+
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设数列a1,a2,…,an,…满足a1=a2=1,a3=2,且对任何自然数n,都有anan+1an+2≠1,又anan+1an+2an+3=an+an+1+an+2+an+3,则a1+a2+…+a100的值是______.
∵对任何自然数n,都有anan+1an+2an+3=an+an+1+an+2+an+3 ①∴an+1an+2an+3an+4=an+1+an+2+an+3+an+4,②
②-①,得anan+1an+2(an+4-an)=an+4-an,即(an+4-an)(anan+1an+2-1)=0
由已知anan+1an+2≠1,即anan+1an+2-1≠0,只能an+4-an=0,即得an+4=an.
又anan+1an+2an+3=an+an+1+an+2+an+3,a1=a2=1,a3=2,得a4=4.
故数列为,1,1,2,4的循环出现
∴a1+a2+…+a100=25(1+1+2+4)=200.
故答案为:200
②-①,得anan+1an+2(an+4-an)=an+4-an,即(an+4-an)(anan+1an+2-1)=0
由已知anan+1an+2≠1,即anan+1an+2-1≠0,只能an+4-an=0,即得an+4=an.
又anan+1an+2an+3=an+an+1+an+2+an+3,a1=a2=1,a3=2,得a4=4.
故数列为,1,1,2,4的循环出现
∴a1+a2+…+a100=25(1+1+2+4)=200.
故答案为:200
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