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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(mǎn)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(mǎn)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(mǎn)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(mǎn)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?