∴an+2=bn+1-2an+1②
②-①×3:an+2-3an+1=(bn+1-3bn)-2an+1+6an
∵bn+1=3bn-4an
∴an+2-3an+1=-2an+1+2an
∴an+2-an+1=2an
∴an+2=bn,即数列{an}从第三项开始与{bn}相同,
∵a1=1,b1=7,∴
a1 |
b1 |
1 |
7 |
设
lim |
n→∞ |
an |
bn |
lim |
n→∞ |
an+1 |
bn+1 |
∴k=
lim |
n→∞ |
bn?2an |
3bn?4an |
1?2k |
3?4k |
∴k=1(舍去)或k=
1 |
4 |
∴
lim |
n→∞ |
an |
bn |
1 |
4 |
故选B.