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=
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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(n 2n+1
)n+1,1 2
则Tn=b1+b2+…+bn
=1•(
)2+2•(1 2
)3+…+(n-1)•(1 2
)n+n•(1 2
)n+1,①1 2
Tn=(1 2
)3+2•(1 2
)4+…+(n-1)•(1 2
)n+1+n•(1 2
)n+2,②1 2
①-②得:
Tn=(1 2
)2+(1 2
)3+…+(1 2
)n+1-n•(1 2
)n+2,1 2
即Tn=
+(1 2
)2+…+(1 2
)n-n•(1 2
)n+1,1 2
∴Tn=
-n•(
(1-(1 2
)n)1 2 1- 1 2
)n+1=1-(1+1 2
)•(n 2
)n.1 2