Question

已知數(shù)列{a_n}中，a1=2，an+1=

an

3an+1

（n為正整數(shù)），依次計(jì)算a₂，a₃，a₄后，歸納、猜想出a_n=

2

6n-5

．

Answer 1

分析：由題意可得，a1=2=

2

1

，a2=

a1

3a1+1

=

2

7

，a3=

a2

3a2+1

=

2

13

=

2

2×6+1

，a4=

a3

3a3+1

=

2

19

=

2

3×6+1

，結(jié)合分母的規(guī)律可猜想

解答：解：由題意可得，a1=2=

2

1

a2=

a1

3a1+1

=

2

7

=

2

6+1

a3=

a2

3a2+1

=

2

13

=

2

2×6+1

a4=

a3

3a3+1

=

2

19

=

2

3×6+1

故猜想，an=

2

6(n-1)+1

=

2

6n-5

故答案為：

2

6n-5

點(diǎn)評：本題主要考查了利用數(shù)列的遞推關(guān)系求解數(shù)列的項(xiàng)，及歸納推理的應(yīng)用．屬于基礎(chǔ)試題