an−1 |
an−1+2 |
anan-1+2an=an-1,
即1+
2 |
an−1 |
1 |
an |
∴
1 |
an |
1 |
an−1 |
∴数列{
1 |
an |
1 |
a1 |
则
1 |
an |
an=
1 |
2n−1 |
由ak>
1 |
2009 |
1 |
2k−1 |
1 |
2009 |
即2k<2010.
∵k为正整数,
∴k的最大值为10.
故选:D.
an−1 |
an−1+2 |
1 |
2009 |
an−1 |
an−1+2 |
2 |
an−1 |
1 |
an |
1 |
an |
1 |
an−1 |
1 |
an |
1 |
a1 |
1 |
an |
1 |
2n−1 |
1 |
2009 |
1 |
2k−1 |
1 |
2009 |