Question

若數(shù)列{a_n}滿足an+1=1-

1

an

，a₁=2，則a₂₀₀₉=

1

2

．

Answer 1

分析：先由an+1=1-

1

an

，可推得數(shù)列的周期，利用數(shù)列的周期性可求得答案．

解答：解：由an+1=1-

1

an

，得
an+3=1-

1

an+2

=1-

1

1- 1 an+1

=1-

an+1

an+1-1

=1-

1- 1 an

1- 1 an -1

=1-

an-1

-1

=a_n，
∴3是數(shù)列{a_n}的周期，
又a2=1-

1

a1

=1-

1

2

=

1

2

，
∴a₂₀₀₉=a_3×669+2=a₂=

1

2

，
故答案為：

1

2

．

點(diǎn)評(píng)：本題考查由數(shù)列遞推式求數(shù)列的項(xiàng)，考查數(shù)列的函數(shù)特性，解決本題的關(guān)鍵是由遞推式推導(dǎo)數(shù)列周期．