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1、專(zhuān)題對(duì)點(diǎn)練14 數(shù)列與數(shù)列不等式的證明及數(shù)列中的存在性問(wèn)題
專(zhuān)題對(duì)點(diǎn)練第19頁(yè)
1.若數(shù)列{an}滿足:a1=23,a2=2,3(an+1-2an+an-1)=2.
(1)證明:數(shù)列{an+1-an}是等差數(shù)列;
(2)求使1a1+1a2+1a3+…+1an>52成立的最小的正整數(shù)n.
(1)證明 由3(an+1-2an+an-1)=2可得
an+1-2an+an-1=23,
即(an+1-an)-(an-an-1)=23,
故數(shù)列{an+1-an}是以a2-a1=43為首項(xiàng),23為公差的等差數(shù)列.
(2)解 由(1)知an+1-an=43+23(n
2、-1)=23(n+1),
于是累加求和得an=a1+23(2+3+…+n)=13n(n+1),
∴1an=31n-1n+1.
∴1a1+1a2+1a3+…+1an=3-3n+1>52,
∴n>5.∴最小的正整數(shù)n為6.
2.(2017廣東揭陽(yáng)二模,理17)已知數(shù)列{an},a1=1,an+1=2(n+1)ann+n+1.
(1)求證:數(shù)列ann+1是等比教列;
(2)求數(shù)列{an}的前n項(xiàng)和Sn.
(1)證明 ∵an+1=2(n+1)ann+n+1,
∴an+1n+1=2×ann+1,∴an+1n+1+1=2×ann+1,
∴數(shù)列ann+1是
3、等比教列,公比為2,首項(xiàng)為2.
(2)解 由(1)可得ann+1=2n,可得an=n·2n-n.
設(shè)數(shù)列{n·2n}的前n項(xiàng)和為T(mén)n,
則Tn=2+2×22+3×23+…+n·2n,
2Tn=22+2×23+…+(n-1)·2n+n·2n+1,
兩式相減可得-Tn=2+22+…+2n-n·2n+1=2(1-2n)1-2-n·2n+1,
∴Tn=(n-1)·2n+1+2.
∴Sn=(n-1)·2n+1+2-n(n+1)2.
3.已知數(shù)列{an}的前n項(xiàng)和Sn=1
4、+λan,其中λ≠0.
(1)證明{an}是等比數(shù)列,并求其通項(xiàng)公式;
(2)若S5=3132,求λ.
解 (1)由題意得a1=S1=1+λa1,故λ≠1,a1=11-λ,a1≠0.
由Sn=1+λan,Sn+1=1+λan+1得an+1=λan+1-λan,
即an+1(λ-1)=λan.
由a1≠0,λ≠0得an≠0,所以an+1an=λλ-1.
因此{(lán)an}是首項(xiàng)為11-λ,公比為λλ-1的等比數(shù)列,
于是an=11-λλλ-1n-1.
(2)由(1)得Sn=1-λλ-1n.由S5=3132得1-λλ-15=3132,即λλ-15=132.解得λ=-1.
4.(201
5、7吉林白山二模,理17)在數(shù)列{an}中,設(shè)f(n)=an,且f(n)滿足f(n+1)-2f(n)=2n(n∈N*),且a1=1.
(1)設(shè)bn=an2n-1,證明數(shù)列{bn}為等差數(shù)列;
(2)求數(shù)列{an}的前n項(xiàng)和Sn.
(1)證明 由已知得an+1=2an+2n,
∴bn+1=an+12n=2an+2n2n=an2n-1+1=bn+1,
∴bn+1-bn=1.又a1=1,∴b1=1,
∴{bn}是首項(xiàng)為1,公差為1的等差數(shù)列.
(2)解 由(1)知,bn=an2n-1=n,∴an=n·2n-1.
∴Sn=1+2×21+3×22+…+n
6、83;2n-1,
2Sn=1×21+2×22+…+(n-1)·2n-1+n·2n,
兩式相減得-Sn=1+21+22+…+2n-1-n·2n=2n-1-n·2n=(1-n)2n-1,
∴Sn=(n-1)·2n+1.
5.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且(3-m)Sn+2man=m+3(n∈N*),其中m為常數(shù),且m≠-3.
(1)求證:{an}是等比數(shù)列;
(2)若數(shù)列{an}的公比q=f(m),數(shù)列{bn}滿足b1=a1,bn=32f(bn-1)(n∈N*,n≥2),求證:1bn為等差數(shù)列,并求bn.
證明
7、 (1)由(3-m)Sn+2man=m+3,
得(3-m)Sn+1+2man+1=m+3,
兩式相減,得(3+m)an+1=2man.
∵m≠-3,∴an+1an=2mm+3,∴{an}是等比數(shù)列.
(2)由(3-m)Sn+2man=m+3,
得(3-m)S1+2ma1=m+3,即a1=1,∴b1=1.
∵數(shù)列{an}的公比q=f(m)=2mm+3,
∴當(dāng)n≥2時(shí),bn=32f(bn-1)=32·2bn-1bn-1+3,
∴bnbn-1+3bn=3bn-1,∴1bn-1bn-1=13.
∴1bn是以1為首項(xiàng),13為公差的等差數(shù)列,
∴1bn=1+n-13=n+23
8、.
又1b1=1也符合,∴bn=3n+2.
6.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=-2,且滿足Sn=12an+1+n+1(n∈N*).
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)若bn=log3(-an+1),求數(shù)列1bnbn+2的前n項(xiàng)和Tn,并求證Tn<34.
(1)解 ∵Sn=12an+1+n+1(n∈N*),∴當(dāng)n=1時(shí),-2=12a2+2,解得a2=-8.
當(dāng)n≥2時(shí),an=Sn-Sn-1=12an+1+n+1-12an+n,
即an+1=3an-2,可得an+1-1=3(an-1).
當(dāng)n=1時(shí),a2-1=3(a1-1)=-9,
∴數(shù)列{an-1}是等比
9、數(shù)列,首項(xiàng)為-3,公比為3.
∴an-1=-3n,即an=1-3n.
(2)證明 bn=log3(-an+1)=n,∴1bnbn+2=121n-1n+2.
∴Tn=121-13+12-14+13-15+…+1n-1-1n+1+1n-1n+2=121+12-1n+1-1n+2<34.∴Tn<34. ?導(dǎo)學(xué)號(hào)16804192?
7.(2017湖南長(zhǎng)郡中學(xué)模擬,理17)在數(shù)列{an}中,Sn為其前n項(xiàng)和,an>0,且4Sn=an2+2an+1(n∈N*),數(shù)列{bn}為等比數(shù)列,公比q>1,b1=a1,且2b2,b4,3b3成等差數(shù)列.
(1)求{an}與{bn}的
10、通項(xiàng)公式;
(2)令cn=anbn,若{cn}的前n項(xiàng)和為T(mén)n,求證:Tn<6.
(1)解 4Sn=an2+2an+1(n∈N*),①
當(dāng)n=1時(shí),4a1=a12+2a1+1,解得a1=1.
當(dāng)n≥2時(shí),4Sn-1=an-12+2an-1+1,②
①-②得4an=(an+1)2-(an-1+1)2,
即(an+an-1)(an-an-1-2)=0.
又an>0,∴an-an-1-2=0,即an-an-1=2,
∴數(shù)列{an}是等差數(shù)列,公差為2.∴an=1+2(n-1)=2n-1.
∵2b2,b4,3b3成等差數(shù)列,
∴2b4=2b2+3b3.∴2b2q2=2b
11、2+3b2q,即2q2-3q-2=0,解得q=2.又b1=a1=1,∴bn=2n-1.
(2)證明 cn=anbn=2n-12n-1,則{cn}的前n項(xiàng)和為T(mén)n=1+32+522+…+2n-12n-1,12Tn=12+322+…+2n-32n-1+2n-12n,
∴12Tn=1+212+122+…+12n-1-2n-12n=1+2×121-12n-11-12-2n-12n,∴Tn=6-2n+32n-1<6. ?導(dǎo)學(xué)號(hào)16804193?
8.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,且對(duì)任意正整數(shù)n,點(diǎn)(an+1,Sn)都在直線2x+y-2=0上.
(1)求數(shù)列{an}
12、的通項(xiàng)公式;
(2)是否存在實(shí)數(shù)λ,使得數(shù)列Sn+λn+λ2n為等差數(shù)列?若存在,求出λ的值;若不存在,請(qǐng)說(shuō)明理由.
解 (1)由題意,得2an+1+Sn-2=0.①
當(dāng)n≥2時(shí),2an+Sn-1-2=0.②
①-②,得2an+1-2an+an=0(n≥2),所以an+1an=12(n≥2).
因?yàn)閍1=1,2a2+a1=2,所以a2=12.
所以{an}是首項(xiàng)為1,公比為12的等比數(shù)列.
所以數(shù)列{an}的通項(xiàng)公式為an=12n-1.
(2)由(1)知,Sn=1-12n1-12=2-12n-1.
若Sn+λn+λ2n為等差數(shù)列,則S1+λ+λ2,S2+2λ+λ22,S3+3λ+λ23成等差數(shù)列,則2S2+9λ4=S1+3λ2+S3+25λ8,
即232+9λ4=1+3λ2+74+25λ8,解得λ=2.
又當(dāng)λ=2時(shí),Sn+2n+22n=2n+2,
顯然{2n+2}是等差數(shù)列.故存在實(shí)數(shù)λ=2,使得數(shù)列Sn+λn+λ2n為等差數(shù)列. ?導(dǎo)學(xué)號(hào)16804194?
高考數(shù)學(xué)理二輪專(zhuān)題復(fù)習(xí)突破精練：專(zhuān)題對(duì)點(diǎn)練14 數(shù)列與數(shù)列不等式的證明及數(shù)列中的存在性問(wèn)題 Word版含解析
