问题
解答题
设数列{an}的前n项和为Sn,a1=1,Sn+1=4an+2(n∈N+) (1)若bn=an+1-2an,求bn; (2)若cn=
(3)若dn=
|
答案
解(1)∵a1=1,Sn+1=4an+2(n∈N+),
∴Sn+2=4an+1+2an+2=Sn+2-Sn+1=4(an+1-an),
∴an+2-2an+1=2(an+1-2an)
即bn+1=2bn
∴{bn}是公比为2的等比数列,且b1=a2-2a1
∵a1=1,a2+a1=S2
即a2+a1=4a1+2,
∴a2=3a1+2=5,
∴b1=5-2=3,
∴bn=3•2n-1.
(2)∵cn=
1 |
an+1-2an |
1 |
bn |
1 |
3•2n-1 |
∴c1=
1 |
3•21-1 |
1 |
3 |
1 |
3 |
1 |
2 |
∴{cn}是首项为
1 |
3 |
1 |
2 |
∴T6=
| ||||
1-
|
2 |
3 |
1 |
64 |
61 |
96 |
(3)∵dn=
an |
2n |
∴dn+1-dn=
an+1 |
2n+1 |
an |
2n |
an+1-2an |
2n+1 |
bn |
2n+1 |
即dn+1-dn=
3•2n-1 |
2n+1 |
3 |
4 |
∴{dn}是等差数列.