一輪復(fù)習(xí)等差等比數(shù)列證明練習(xí)題

1、111 .已知數(shù)列an是首項(xiàng)為ai=-,公比q =的等比數(shù)列，bn+2 =3log1 an(n W N*),數(shù)列cn滿足 Cn=an 'bn .(1)求證：是等差數(shù)列;2,數(shù)列l(wèi)a滿足a1 =2，an書=a2 6an 6( n N )設(shè) Cn =log5(an 3)C(I)求證：cn是等比數(shù)歹U；)3.設(shè)數(shù)列an 的前 n 項(xiàng)和為 Sn,已知 a1 + 2a2 +3a3+nan = (n 1)Sn + 2n (n w N(2)求證：數(shù)列+2是等比數(shù)列;2n 1a4-數(shù)列an滿足 a1 =1, an 書=n-(n 乏 N 4an 2(1)證明：數(shù)列2-是等差數(shù)列; an,2S25 .數(shù)列

2、an 首項(xiàng)a1 =1,前n項(xiàng)和Sn與an之間滿足an = 一(0之2)2Sn-11(1)求證：數(shù)列'卜是等差數(shù)列 Sn26 .數(shù)列 an滿足 a1 = 3 , an +=，an 1a -1 (1)求證：a一成等比數(shù)列； an 2* .7 .已知數(shù)列an滿足 an+ =3an +4, (n = N )且 a = 1 ,(I )求證：數(shù)列an +2是等比數(shù)列；8 .數(shù)列 an滿足：ai =1, n an甲=(n 十 1) an 十 n (n + 1),n w N(1)證明：數(shù)列生是等差數(shù)列；n22a9.已知數(shù)列an的首項(xiàng) a1= , an4= , n=1, 2,31 an 11(1)證明：

3、數(shù)列11、是等比數(shù)列；an10 .已知數(shù)列an的前 n項(xiàng)和為 Sn, a1 =1，Sn =n2an_n(n_1),n=1,2,L2(1)證明：數(shù)列1口二l是等差數(shù)列，并求 Sn；,n11 . (16分)已知數(shù)列an的前n項(xiàng)和是Sn ,且Sn =2an n(1)證明:A +仆為等比數(shù)列;12 .數(shù)列an滿足：a1 =2,a2 =3,an42 = 3an書2an(nw N*)(1)記dn =an+an，求證：數(shù)列dn是等比數(shù)歹U；13 .已知數(shù)列an的相鄰兩項(xiàng)an , an書是關(guān)于X方程x2 -2nx + bn =0的兩根，且a1 =1 .1 C(1)求證：數(shù)列an 2n是等比數(shù)列；314 .(本

4、題滿分12分)已知數(shù)列an中，a1 =5且an =2an+2n 1 (n至2且nW N *).a 1(I)證明：數(shù)列 Jan為等差數(shù)列；,2n15 .已知數(shù)列 I 中，a1 =1,an書=ar(n w N ) an 3r三(1)求證：，工十1 是等比數(shù)列，并求Q的通項(xiàng)公式an;©n 2,35-16.設(shè)數(shù)列an的刖n項(xiàng)和為Sn, n = N ,已知a1 = 1, a2 =，a3 = 一，且當(dāng)n22時(shí),2445七 +5& =8& 書 +&二(1)求a4的值;1(2)證明：Jan+_lan '為等比數(shù)列; n 12 n17 .設(shè)數(shù)列an的前n項(xiàng)和為Sn ,且

5、首項(xiàng)a1 #3,20中=Sn +3n(nw N*).(i)求證：Sn 3n是等比數(shù)歹U;18 .(本小題滿分10分)已知數(shù)列小滿足&=1, a(3n+3)an+4"6” n n(1)求證：數(shù)列但上2!是等比數(shù)列； ,n參考答案1.(1)見解析；(2) Sn =2_(3n +2)n3)m* 或 mW53342.(I)見解析;()丁 11C 一 二2n - QTn - “2n 一an -53. ；(出)4 5-93. (1) a2 =4, a3 =8 ； (2)見解析；(3) 52.4. (1)詳見解析；(2) an=；(3) (2n3)2n +6 n 11n = 1)5. (1

6、)詳見解析;，an=$2,>6；-(n 2)(2n-1)(2n-3)A 1 6. (1)證明a4成等比數(shù)列的過程詳見試題解析;an 2（2）實(shí)數(shù)t的取值范圍為 上Y3 <t w Y3二17. 詳見解析a2n -1 3n 138. (1)見解析；(2) Sn 49. (1)詳見解析(2) s =2-+ n(n+1)n J _nd n22210. . ( 1 )由Sn=n2an n(n1)知，當(dāng) n 22 時(shí)，Sn=n2(S S1) nA1)即22n - 1 n1 1(n 1)Sn n Sn= n(n 1),所以Sn -Sn= 1,對(duì) nA2 成立.又S1=1, nn 11n 1rSn

7、 =1 (n - 1) 1,即 n所以【DUsnl是首項(xiàng)為 1,公差為 1的等差數(shù)列.所以 ,n因 為bn = 3 sn 2 =1= 1(-), 所 以n 3n (n 1)(n 3) 2 n 1 n 3bi b2 L-)=1(- n 32 611 、5):.n 2 n 312k=18 k=6 k=411. (1)見解析；(2)解析；(3)存在， 或W 或士m=18 m=5 m = 212. (1) dn =1M2n,(2) an =2n+1r9n 由 2-n為偶數(shù)13. (1)見解析；(2) Sn = 34t 3，(3)(-°0,1)-1 n為奇數(shù),3314. ( I )詳見解析(n)& =n 2n