n1
(2)由(1)得,cn=n-,
2?n+1??n+2?
n??1?12
故Tn=c1+c2+…+cn=?++…+n?-?2??233?24
+
1
+…+334
1?.
??n+1??n+2??
12n112n1111n设Fn=++…+n,则Fn=2+3+…+n+1,作差得Fn=+2+…+n-n+1,
242222222222所以Fn=2-n+2
2
n. 11111111111
设Gn=++…+=-+-+…+-=-,所以
233334?n+1??n+2?2334n+1n+22n+21?3n+2n+2?11
Tn=2-n-?-=-n+. ?2n+2?2n+2?22
17.(20172山东高考)已知{an}是各项均为正数的等比数列,且a1+a2=6,a1a2=a3. (1)求数列{an}的通项公式;
(2){bn}为各项非零的等差数列,其前n项和为Sn.已知S2n+1=bnbn+1,求数列??的前n?an??bn?
项和Tn.
解 (1)设{an}的公比为q, 由题意知a1(1+q)=6,a1q=a1q,
又an>0,由以上两式联立方程组解得a1=2,q=2, 所以an=2. (2)由题意知
?2n+1??b1+b2n+1?
=(2n+1)bn+1,
2
n2
2
S2n+1=
又S2n+1=bnbn+1,bn+1≠0, 所以bn=2n+1. 令cn=,则cn=
bnan2n+1
. n2
因此Tn=c1+c2+…+cn
3572n-12n+1=+2+3+…+n-1+n, 2222213572n-12n+1又Tn=2+3+4+…+n+n+1, 222222两式相减得
1?2n+113?11
Tn=+?+2+…+n-1?-n+1,
2?222?222n+5
所以Tn=5-n.
2
18.在等比数列{an}中,a1>0,n∈N,且a3-a2=8,又a1,a5的等比中项为16. (1)求数列{an}的通项公式;
*
111
(2)设bn=log4an,数列{bn}的前n项和为Sn,是否存在正整数k,使得+++…+S1S2S3
1
Sn * a3-a2=8,则a2=8,q=2,a1=4,所以an=2n+1. (2)bn=log42 n+1 = n+1 2 , . Sn=b1+b2+…+bn= 1=n?n+3? 4 Sn1?44?1 =?-?, n?n+3?3?nn+3? 1111所以+++…+ S1S2S3Sn11?4?111111 =?-+-+-+…+- nn+3?3?142536?111?4?11 --=?1++-? 3?23n+1n+2n+3?11?4114?1++=3-3?? 363?n+1n+2n+3?11?224?1++=-3??. 93?n+1n+2n+3?122 当n=1时,=1<2<; S19 11?22111224?1 ++当n≥2时,++…+=-??<<3. S1S2Sn93?n+1n+2n+3?9故存在k=3时,对任意的n∈N都有 1 * S1S2S3 111 +++…+<3. Sn
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