(2013•日照二模)设数列{an}的各项都是正数,且对任意n∈N*,都有(an-1)(an+3)=4Sn,其中Sn为数
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(2013•日照二模)设数列{an}的各项都是正数,且对任意n∈N*,都有(an-1)(an+3)=4Sn,其中Sn为数列{an}的前n项和.
(Ⅰ)求证数列{an}是等差数列;
(Ⅱ)若数列{
}
(Ⅰ)求证数列{an}是等差数列;
(Ⅱ)若数列{
4 | ||
|
(Ⅰ)∵(an-1)(an+3)=4Sn,当n≥2时,(an-1-1)(an-1+3)=4Sn-1,
两式相减,得
a2n−
a2n−1+2an−2an−1=4an,即(an+an-1)(an-an-1-2)=0,又an>0,∴an-an-1=2.
当n=1时,(a1-1)(a1+3)=4a1,∴(a1+1)(a1-3)=0,又a1>0,∴a1=3.
所以,数是以3为首项,2为公差的等差数列.
(Ⅱ)由(Ⅰ),a1=3,d=2,∴an=2n+1.
设bn=
4
an2−1,n∈N*;∵an=2n+1,∴an2−1=4n(n+1)))
∴bn=
4
4n(n+1)=
1
n(n+1)=
1
n−
1
n+1
∴Tn=b1+b2+b3+…+bn=(1−
1
2)+(
1
2−
1
3)+…+(
1
n−
1
n+1)=1−
1
n+1<1.
又∵Tn+1−Tn=
n+1
n+2−
n
n+1=
1
(n+2)(n+1)>0,∴Tn+1>Tn>Tn−1>…>T1=
1
2,
综上所述:不等式
1
2≤Tn<1成立.
两式相减,得
a2n−
a2n−1+2an−2an−1=4an,即(an+an-1)(an-an-1-2)=0,又an>0,∴an-an-1=2.
当n=1时,(a1-1)(a1+3)=4a1,∴(a1+1)(a1-3)=0,又a1>0,∴a1=3.
所以,数是以3为首项,2为公差的等差数列.
(Ⅱ)由(Ⅰ),a1=3,d=2,∴an=2n+1.
设bn=
4
an2−1,n∈N*;∵an=2n+1,∴an2−1=4n(n+1)))
∴bn=
4
4n(n+1)=
1
n(n+1)=
1
n−
1
n+1
∴Tn=b1+b2+b3+…+bn=(1−
1
2)+(
1
2−
1
3)+…+(
1
n−
1
n+1)=1−
1
n+1<1.
又∵Tn+1−Tn=
n+1
n+2−
n
n+1=
1
(n+2)(n+1)>0,∴Tn+1>Tn>Tn−1>…>T1=
1
2,
综上所述:不等式
1
2≤Tn<1成立.
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