Question

设数列{an}满足：a1=a2=1，a3=2，且对于任意正整数n都有anan+1an+2≠1，又anan+1an+2an+3=an+an+1+an+2+a

设数列{an}满足：a1=a2=1，a3=2，且对于任意正整数n都有anan+1an+2≠1，又anan+1an+2an+3=an+an+1+an+2+an+3，则a1+a2+a3+…+a2013=______．

Answer 1

a₁=a₂=1，a₃=2，
又∵a_na_n+1a_n+2≠1，且a_na_n+1a_n+2a_n+3=a_n+a_n+1+a_n+2+a_n+3，
则an+3＝

an+an+1+an+2

an?an+1?an+2?1

∴a₄=

1+1+2

2?1

=4
a₅=

1+2+4

1×2×4?1

=1
a₆=

2+4+1

1×2×4?1

=1
a7＝

4+1+1

4×1×1?1

=2
a₈=

1+1+2

1×1×2?1

=4
由以上可发现数列{a_n}是以4为周期的数列
则a₁+a₂+a₃+…+a₂₀₁₃=503（a₁+a₂+a₃+a₄）+a₁
=503×8+1=4025
故答案为：4025