问题
解答题
已知数列{an}满足an+1=
(1)令bn=
(2)求满足am+am+1+…+a2m-1<
|
答案
(1)由an+1=
an |
3-2an |
1 |
an+1 |
3 |
an |
∴
1 |
an+1 |
1 |
an |
∵
1 |
a1 |
1 | ||
|
∴数列{
1 |
an |
∴
1 |
an |
bn=3n(n∈N*).
(20由(1)可知:an=
1 |
3n+1 |
∴am+am+1+…+a2m-1=
1 |
3m+1 |
1 |
3m+1+1 |
1 |
32m-1+1 |
<
1 |
3m |
1 |
3m+1 |
1 |
32m-1 |
1 |
3m |
1-
| ||
1-
|
1 |
2×3m-1 |
1 |
3m |
<
1 |
2×3m-1 |
令
1 |
2×3m-1 |
1 |
150 |
故所求m的最小值为5.