完整版數(shù)列證明題型總結教師版附答案

1、、解答題1 -在數(shù)列an中，a1 = 1, an+1 = 2an + 2n.an(I )設bn= 尹，證明：數(shù)列bn是等差數(shù)列;(n )求數(shù)列an的前n項的和Sn.【答案】(I)因為an+1 an an+1 2an 2n刁=12n所以數(shù)列bn為等差數(shù)列(n)因為bn= bi + (n 1) 1 = n5 / 18所以 an= n 2n 2所以 Sn= 1 怎0+ 2X21 + + nn12Sn = 1 k1 + 2X22+ + nX2n兩式相減得Sn= (n 1) 2n+ 12 -在數(shù)列an中，a1 = 2， an + 1= 2an + 2n+ 1.(I )設bn = 2nan，證明：數(shù)列bn

2、是等差數(shù)列;(n )求數(shù)列an的前n項和Si.【答案】丄11(I )由 an+1 = 2an + 2*+1,得 2n 1an + 1 = 2nan + 1 bn+1= bn+ 1 ,則bn是首項b1= 1，公差為1的等差數(shù)列.故 bn= n, an =藥.(n )Sn= 1 X2 +2X*+ 33+ (n- 1)+n X1!|sn = 134 + (n 1) 2i+兩式相減，得:|sn = 2 +寺+23+寺一芋n_= 1 112n+ 12n 2n+1S 2 丄一 nSn= 2 2*-1 2*3.數(shù)列an的各項均為正數(shù)，前 n項和為Sn,且滿足4Sn= (an+ 1)2(n N*).(I )證

3、明：數(shù)列an是等差數(shù)列，并求出其通項公式an； (n )設 bn = an + 2an(n N )，求數(shù)列bn的前 n 項和 Tn.【答案】(I )n = 1 時，4a1 = (a1 + 1)2? a2 2a1 + 1 = 0，即 a1 = 1 n時，4an = 4Sn 4Sn-1 = (an + 1)2 (an-1 + 1)2= an an-1 + 2an 2an- 1? an一 an 1 一 2an一 2an1 = 0 ? (an + an T)( an an-1) 2 = 0-an0 二 an an1= 2 故數(shù)列an是首項為a1= 1,公差為d = 2的等差數(shù)列，且 an= 2n 1(

4、n N*) (n )由(I )知 bn = an + 2an= (2n 1) + 22n1 -Tn= b1 + b2 + + bn =(1 + 21)+ (3 + 23) + (2 n 1) + 22n-1 =1 + 3+ + (2n 1) + (21 + 23+ 22n 1)2 2 ( 1 22n)22n+12 2 22n+1+ 302 2=卡+1 4= 丁 + 宀 2=4. 數(shù)列an的各項均為正數(shù)，前n項和為Sn,且滿足2/Si= an+1(n N*).(I )證明：數(shù)列an是等差數(shù)列，并求出其通項公式an； (n )設 bn = an 2n(n N*)，求數(shù)列bn的前 n 項和 Tn.【

5、答案】(I )由 VSn= an+ 1(n N )可以得到 4Sn= (an+ 1)2(n N ) n= 1 時，4a1 = (a1 +1)2? a? 2a1 + 1 = 0，即 a1= 1 n2時，4an = 4Si 4Sn1 = (an+ 1)2 (an 1 + 1)2 =ain a2 + 2an 2an1 ? an aH-1 2an 2an-1 = 0 ? (an + an-1)( an an-1) 2 = 0-an0- an an 1 = 2 故數(shù)列an是首項為a1 = 1,公差為d = 2的等差數(shù)列，且 an= 2n 1(n N*)(n )由( I )知 bn = an 2n= (2

6、n 1) 2n 二Tn= (1 21)+ (3 22) + (2n 3) 2nJ + (2n 1) 2n 則 2Tn= (1 22) + (3 23) + (2 n 3) 2n + (2 n 1) 2n+1 兩式相減得: Tn= (1 21)+ (2 22)+ + (2 2n) (2n 1) 2n+12 / 1 2n)=2 2(2 (2n 1) 2n+1 =(3 2n )2n+1 6 二 Tn= (2n 3) 2n + 6(或 Tn= (4n 6) 2n + 6)37*5. 已知數(shù)列an，其前n項和為Sn=+ 2n(n N ).(I )求 ai, a2;(n )求數(shù)列an的通項公式，并證明數(shù)列

7、 an是等差數(shù)列;(川)如果數(shù)列bn滿足an= log2bn，請證明數(shù)列bn是等比數(shù)列，并求其前 n項和Tn.【答案】(I )ai= Si= 5, a1 + a2= S2= |x22 + 7 X2 = 13, 解得a2= 8.(n )當 n2時, an= Sn Sn 1=|n2 (n 1)2 + 7n (n 1) =2(2n 1) + 2= 3n+ 2.又 ai = 5 滿足 an = 3n + 2, 二 an= 3n + 2(n N*).an an-1= 3ri+ 2 3(n 1)+ 2 =3(n2 n N*), 數(shù)列an是以5為首項，3為公差的等差數(shù)列.(川)由已知得bn= 2an(n N

8、*),bpnn +1bn2an心=2an+1 an= 23= 8(n N*)，又 bi = 2a1= 32, 數(shù)列bn是以32為首項，8為公比的等比數(shù)列.Tn= 38 = 32(8n 1).2x46. 已知函數(shù) f(X)= x+ 2，數(shù)列an滿足：a1 = 3， an + 1= f(an).an的通項公式;1(I )求證：數(shù)列一為等差數(shù)列，并求數(shù)列an求證:(n 己 Sn = aia2 + a2a3 + + anan+1,【答案】2an證明：d ) an+1 = f(an)= an+ 2,=丄+1,an+1 an 21 1 1 即an+1 an則a成等差數(shù)列,所以 1 = 1 + (n 1)

9、X2 = 4+ (n 1)ana1an =2n + 1(n 廠anan+1=2+7 齊=82n+12n+，1 一 1 1 8 =8 3 2n + 3 習.1 1+ 1 1丄丄Sn= a1a2 + a2a3 + anan+1 = 8 3 5+ 5 7+ + 2n+ 1 2ri+ 37. 已知數(shù)列an的前三項依次為2, 8, 24，且an 2an-1是等比數(shù)列.(I)證明器是等差數(shù)列;(n )試求數(shù)列an的前n項和Sn的公式.【答案】(I ) a2 2a1 = 4, a3 2a2= 8, -an 2an-1是以2為公比的等比數(shù)列.an 2anT = 4 2“ 2= 21 等式兩邊同除以2n,得2

10、齊=1 ,an是等差數(shù)列.(n )根據(jù)(I )可知 2= y+(n 1) X = n,.an= n 2n.Sn= 1 X2+ 2X22+ 323 + n 2n, 2Sn = 1 X22+ 2X23+ + (n 1) 2n+n 2n+1. 一得:Sn= 2+ 22+ 23+ + 2n n 2n+11 2=2(1 2) n 2n+1= 2n+1 2 n 2n+1 二 Sn= (n 1) 2n+1+ 2.1 18. 已知數(shù)列an的各項為正數(shù)，前 n項和為Sn,且滿足：Sn = - an+_ (n N*).2 an(I)證明：數(shù)列Sn是等差數(shù)列;(n )設 Tn = *&+ 212S2+ 23s2+

11、2&，求 Tn.【答案】11(I )證明：當 n = 1 時，a1= S1,又 Sn = 2 an + (n N*),11二 S1 = 1 S1 + &，解得 S1 = 1.2S1當 n2 時，an= Sn Sn-1 ,1 Sn = 1 S-$- 1 + Sn s 1，1即 S+S-1=Sn，化簡得 s2-s2-1=1,Sn是以S1= 1為首項，1為公差的等差數(shù)列.(n )由(I )知 Sn = n,Tn=G+Is2* + ns2,即 Tn= 1 1+ 2 22+ (n 1+ n 如11 1 1 1夯得 1Tn= 1 護+ (n 1)21n + n +1 . 得如=1+右+寺-n 121-寺-

12、n十=1-芬+= 1 十2*+1，-Tn= 22n9.數(shù)列an滿足 a1 = 1, an+1+ 4= 1(n N*)，記 Sn= a2+ a2+ + a2.1(I )證明：72是等差數(shù)列;an(n )對任意的n N*，如果S2n+1 Sn詰0恒成立，求正整數(shù) m的最小值.【答案】11 11 1)證明：017一02=4?晶=01+(n一1)関？懇=4n 3,1冷是等差數(shù)列. an1 1 1)令 g(n)=9n+1-Sn=時+8 g(n+ 1) - g(n )n1 = 3X2n 則 an+1 an= bn2 = 3 2n 2 , 所以 an= a1+ (a2 a1) + (a3 a2) + + (

13、an an1) =1 + (3 21 2) + (3 22 2) + (3 2n1 2) =1 + 3(2 + 22+ 23+ 2n 1) 2(n 1) an = 3 2n 2n 3, 當n= 1時，a1 = 3怎1 2X1 3 = 6 5= 1，故a1也滿足上式 故數(shù)列an的通項為an= 3 Q 2n 3(n N*).12.在數(shù)列 an中，a1 = 6, an= 1an-1 +n N*且 n2) (I )證明：an+耳是等比數(shù)列;(n )求數(shù)列an的通項公式;(川)設Sn為數(shù)列 an的前n項和，求證SiV?.【答案】(I)由已知，得/11丄、丄1an + 1 + 3n+ 1( 2an + 2

14、 3n+ 1)十 3n+ 11an+ 3*1an+ 3*=2- an+ 3是等比數(shù)列.9 / 18q=11 111(n )設 An= an+3n,貝y A1 = a1 + 1 = 6十3= 2，且 則 An= (2)n ,11 十曰11-an+ 3n=歹，可得an=尹一羅(n )Sn= & 事)+ + &-1)11,1112 3n 2n 1i L2 32 2nV-n22 6n 、213.已知數(shù)列an滿足 a1 = 2, an+1 = 2an n+ 1(n N ).(I )證明：數(shù)列an n是等比數(shù)列，并求出數(shù)列 an的通項公式;(H)數(shù)列bn滿足：bn=2an(n N*)，求數(shù)列bn的前n項和

15、Sn.【答案】(I )證法一：由 an+1 = 2an一 n + 1 可得 an+1 一(n+ 1) = 2(an一 n),又 a1 = 2，貝U a1 一 1 = 1, 數(shù)列an n是以ai 1 = 1為首項，且公比為 2的等比數(shù)列,則 an n= 1 n1, an= 2n1 + n.證法二:an +1 ( n + 1)2an n+ 1 一 ( n+ 1)2an 2n=2,an nan nan n又 a1 = 2,貝y a1 1 = 1,數(shù)列an n是以ai 1 = 1為首項，且公比為2的等比數(shù)列,則 an n= 1 n1, an= 2n1 + n.(吐廠 bn= 20bn=207=刁-Sn

16、= b1+b2+ bn= 2 + 2(2)2 + n (訓11 1 1 1 2Sn= (了 + 2(2)3 + + (n- 1)(了 + n q111 1 1由一，得 2Sn= 1 + q2 + (p3 + (pn 切-(1)n1n(2)n+1= 1 (n +2)(捫1, Sn= 2 (n+ 2)(2)n.(I )設bn = an證明：數(shù)列bn是等比數(shù)列;(n )求數(shù)列an的前n項和Sn.【答案】bn+1an+1n 1bn n+1 an2(I )因為L=-所以bn是首項為1,公比為2的等比數(shù)列(n )由(I)可知罟二，即 an =,2,3,4$ = 1 + 2 + 22+ 23+ + 2n-1

17、,1上式兩邊乘以M得112,3, n 1 , n2$=2+22+藝+ 莎?+刁1 1 1 1兩式相減，得2Sn=丨+孑+尹+ 2n-1- 2,所以Sn= 4-學n15.設數(shù)列an的前n項和為Sn,且Sn= (1 +為一入0,其中心1 , 0.(I )證明：數(shù)列an是等比數(shù)列;1(n )設數(shù)列an的公比 q= f(?),數(shù)列bn滿足 b1 = 2, bn= f(bn-1)(n N*, n2)求數(shù)列bn的通項公式.【答案】(I )由 Sn= (1 + 入)-入 an Sn 1 = (1 + 入)-入 am 1(n2)an相減得:an=-X al a花=十0 2)數(shù)列an是等比數(shù)列bn 1 ,(n

18、)f( =)+；, bn=1770二？b0=bn；+1bn 占是首項為21= 2,公差為1的等差數(shù)列;右=2+ (n - 1) = n+ 1 , bn= bn ，n+116.在等差數(shù)列an中，a10 = 30, a20= 50.(I )求數(shù)列an的通項an； (n )令bn = 2an 10，證明：數(shù)列bn為等比數(shù)列;(川)求數(shù)列nbn的前n項和Tn.【答案】(I )由 an = a1 + (n 1)d, a10= 30, a20 = 50, 得方程組 a1+ 9d = 30，解得 a1= 12, d= 2.a1+ 19d= 50二 an= 12+ (n 1) 2 = 2n+ 10.b 4n

19、+1(n )由(I )得 bn = 2an 10 = 22n+1010 = 22n= 4n,A 十=廠=4二bn是首項是4,公比q= 4的等比數(shù)列.(川)由 n bn= n X4n 得：Tn= 1 用 + 24n+ n X4n1 相減可得:23 / 183Tn= 4 +42+ + 4n n X4n + 1 = 4( 14n)3n X4n1(3n 1)總n+1 + 4 Tn =17.已知an是等差數(shù)列，其前n 項和為 Si，已知 a3= 11, S9= 153, (I )求數(shù)列an的通項公式;(n )設an= Iog2bn，證明 bn是等比數(shù)列，并求其前 n項和Tn.【答案】a1 + 2d =

20、11(1 ) 9a1 + 爭=153 解得：d= 3, a1 = 5, - an= 3n + 2b + 12 an+ 1(n )bn= 2an,T 詈=缶=2an+1 an= 23= 8,bn2二bn是公比為8的等比數(shù)列又 b1 = 2a1= 32, Tn= 32 (1-)= (刃一1).(I )求a2, a3的值;(n )證明：數(shù)列an + n是等比數(shù)列，并求an的通項公式;(川)求數(shù)列an的前n項和S1.【答案】(I )T ai = 3, an= 2an-1 + n 2(n2 且 n N*), 二 a2= 2ai + 2 2 = 6,an+ na3= 2a2 + 3 2= 13.(n)證明

21、： 廠-們=(細-1+-21 + n an1 +( n 1)an1 + n 1=2an 1+ 2n2 = 2, an 1 + n 1數(shù)列an + n是首項為ai+ 1 = 4，公比為2的等比數(shù)列. an+n= 4 2n勺二2n+ 1,即卩 an = 2n +1 n, 二 an的通項公式為 an= 2n 1 n(n N ).(川廠an的通項公式為an= 2n+1 n(n N*), Sn= (22 + 23 + 24+ 2n+1)(1 +2+ 3 + + n) =22X (1 2n) n X ( n+ 1) =2n+2-19.已知數(shù)列an滿足 a1 = 2, an+1 = 3an+ 2(n N*)

22、.(I )求證：數(shù)列an+ 1是等比數(shù)列;(n )求數(shù)列an的通項公式.【答案】(I )證明：由 an+1 = 3an + 2 得 an+1 + 1 = 3(an + 1), 從而a = 3,an+ 1 即數(shù)列an + 1是首項為3,公比為3的等比數(shù)列.(n )由(I )知, an + 1 = 3 3n1 = 3n? an= 3n 1.20.已知數(shù)列an滿足a1 = 2, an+1 = 4an+ 21, Sn為an的前n項和. (I )設bn = an + 2n，證明數(shù)列bn是等比數(shù)列，并求數(shù)列an的通項公式;(n)設 Tn=Sn，n 3n= 1, 2, 3,，證明：Ti3. i= 12【答案

23、】(I )因為 bn+ 1 = an + 1 + 21 = (4an + 21)+ 21 = 4(an + 2n)= 4bn，且 b1= a1+ 2 = 4,所以bn是以4為首項，以q = 4為公比的等比數(shù)列.所以 bn= b1qn1 = 4n，所以 an= 4n 2n(n )Sn= ai+a2+ + an= (4+42+ + 4n) (2 + 22+ + 2n)=|(4n 1) 2(2n 1) = 1(2n+1)2 3 2n +1 + 2=3(2n+1 1)(2n+1 2) = |(2n+1 1)(2n 1),2n 3所以 Tn= Sn= 2X( 2n+1 1)( 2n 1) = 2 X 2

24、n 1 22nn+ 1n3因此Ti = 22二7 2n + 1 1i = 12i = 121 1 2n+1 1 2)當n= 1時也滿足，所以數(shù)列bn的通項公式為bn= 3廣1 1.22.在各項均為負數(shù)的數(shù)列an中，已知點(an, an+ 1)(n N*)在函數(shù)y= 3X的圖象上，且a2 a5=8=27.(I )求證：數(shù)列an是等比數(shù)列，并求出其通項;(n )若數(shù)列bn的前n項和為Si,且bn = an+n,求Si.【答案】* 2(I )證明：因為點(an, an+1)(n N)在函數(shù)y =衣的圖象上,所以an+1 = Ian,即= 2,故數(shù)列an是公比q= 2的等比數(shù)列.3an 338 8 2

25、 5 2 3因為 a2a5= 27,貝y a1q a1q4 =習，即卩 a1 3 = 3 ,3由于數(shù)列an的各項均為負數(shù)，貝Ua1 = 2,所以an= 2門一 22門一 2an = 一 3, bn= 3+ n,n 1 n2+n 9+2.(n)由(I)知,2 所以Sn= 3 !23 .已知數(shù)列an的前n項和為Sn,且Sn= 3 2n 1 2 , bn= an + 1.(I )求數(shù)列an的通項公式;(n )證明：數(shù)列bn是等比數(shù)列，并求其前 n項和Tn.【答案】(I ) Sn = 3 2n 1 2,當 n2時，an= Sn Sn-1= 3 2n 1 2 3 n 2+ 2= 3 2n21當n= 1時

26、，a1 = 1不滿足上式. * 3 2n-2(n= 1),(n2).(n )bn= an+1= 3-2= 2, n N* ,數(shù)列bn是首項為bn1 33，公比為2的等比數(shù)列;由等比數(shù)列前n項和公式得Tn= 3(于=3 3.24.設數(shù)列an的前 n 項和為 Sn,已知 a1 = 5, an+1 = &+ 3n(n N*). (I )令 bn = Sn 3n，求證：bn是等比數(shù)列；、 2 011(n )令cn= Iog2bn+1 log2bn + 2設是數(shù)列Cn的前n項和，求滿足不等式 Tn4026的門的最小值.【答案】(I )證明：bi = Si 3= 2工0Sn+ 1 Sn= Sn+ 3“，即

27、卩 Sn+ 1= 2Sn+ 3“ ,bn+1 = Sn+1 3n+1 = 2Sn 3+1+ 3n = 2_0 bn = Sn 3n =-n =0Sn 3n所以bn是等比數(shù)列.(n )由(I )知 bn = 2n ,貝廿Cn =log2bn+1 log2bn+2(n+ 1) (n+ 2)n+ 1 n+ 2112 011c刖Tn =廠衛(wèi)礦，n2 011，即 nmin = 2 012.25.已知數(shù)列an滿足：a1 = 1, an+1 =玄 + ?(n N ).(I)求證：數(shù)列an+1是等比數(shù)列;1(n)若n= -+1,且數(shù)列bn是單調(diào)遞增數(shù)列，求實數(shù) 入的取值范圍.bn+1【答案】1211(I )證

28、明：=1+ -,丄 + 1= 2 + 1 ,ccanan + 1anan+11+ 1 = 2mq所以數(shù)列1+ 1是等比數(shù)列.a1an(n )a+1=卯，an= 2n-1，1n-r匸 1 = 2n, bn+1 = 2n(n-A,bn+ 1bn = 2“ 1(n 1 ?)(n2) b1 =入適合,所以 bn= 2n 1(n-1 - Mn N*),由 bn+1bn得 21(n+ 1 為2n(n- ?),?n + 2,?( n+ 2)min= 3,入的取值范圍為入?2 010的n的最小值.【答案】(I )an+1 = 3an 2an-1 (n2,)(an+1 an)= 2(an an- 1)(n 2.

29、) a1 = 2, a2= 4， - a2 a1 = 2mQ an an-10,故數(shù)列an+1-an是首項為2,公比為2的等比數(shù)列,an+1 an= (a2 a1)2n 1= 2n,.an= (an an 1)+ (an-1 an 2)+ (an 2 an 3)+ +(a2 ai)+ ai=2n1 + 2n2+ 2n3+ 21 + 2=2 (1 - 2n1)+ 2=2n(nA 2.)又 a1 = 2 滿足上式，an = 2n(n N ).,-2 (an 1) c1c1an(n )由(I )知 bn = = 2 1 - = 2 1 歹1 1 1 Sn= 2n 1+尹+尹+ 21C 1戶 111.

30、=2n 1 = 2n 2 1 莎=2n2 +1 1由 Sn2 010 得:2n2+2 010，即 n+ 21 006,因為n為正整數(shù)，所以n的最小值為1 006.27.已知數(shù)列an的前n項和為S,滿足Si+2n=2an.(I )證明：數(shù)列a n+2是等比數(shù)列，并求數(shù)列an的通項公式an；1(n)若數(shù)列bn滿足bn=log2 ( an+2 ),求數(shù)列一 的前n項和Tn. bn【答案】(I)證明：由 Sn+2n=2an，得 Sn=2an- 2n,當 nW N 時，Sn=2an- 2n,當 n=1 時，S1=2a1 - 2，則 a1=2 ,當 n2時，Sn-1=2an-1-2 (n- 1),-，得

31、an=2an - 2an-1 - 2,即 an=2an - 1 +2 ,-an+2=2 (an-1+2),an+2=2 , an 1 +2二a n+2是以ai+2為首項，以2為公比的等比數(shù)列. an +2=4c2n 1- an=2n+12 .(n)解： an=2n+12,- bn=n (n+1),1 = 1 = 1bn n n+1 n1n+1-Tn = + + Lb21+ bn1-+3,1 1=1 + 2 2=1 -丄 n+11 1+ 一n n+1nn+1【解析】考點:數(shù)列的求和；等比數(shù)列的通項公式.專題:綜合題.分析:(I)由 Sn+2n=2an，得Sn=2an 2n,由此利用構造法能夠證明數(shù)列an+2是等比數(shù)列，并求出數(shù)列an的通項公式an.1(n)