问题
解答题
已知递增的等比数列{an}的前三项之积为512,且这三项分别依次减去1、3、9后又成等差数列. (1)求数列{an}的通项公式; (2)若Tn=
|
答案
(1)设递增的等比数列{an}的前三项分别为a1,a2,a3,
则a1a2a3=512,∴a2=8.
又这三项分别依次减去1、3、9后又成等差数列,
则2(a2-3)=a1-1+a3-9,即a1+a3=20.
又∵a1a3=a22=64,且a1<a3,∴a1=4,a3=16,
∴等比数列{an}的公比q=2.
∴an=a1qn-1=4•2n-1=2n+1;
(2)证明:令bn=
n |
an |
n |
2n+1 |
1 |
2 |
则Tn=b1+b2+…+bn
=1•(
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
①-②得:
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
即Tn=
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
∴Tn=
| ||||
1-
|
1 |
2 |
n |
2 |
1 |
2 |