问题 填空题

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______.

答案

∵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

选择题
选择题