1、一轮复习大题专练一轮复习大题专练 32数列(证明不等式问题)数列(证明不等式问题)1已知数列na满足12a ,1122nnnaa(1)证明数列2nna为等差数列;(2)设2nnnab ,证明:1 22311111nnbbb bb b证明: (1)根据题意,12a ,1122nnnaa,在等式左右两边同时除以12n得,11122nnnnaa11122nnnnaa,由此可得,数列2nna是首项为112a,公差为 1 的等差数列(2)由(1)可得,1(1)2nnnabnn 11111(1)1nnb bn nnn,1 223111111111111(1)()()()112233411nnbbb bb
2、bnnn 从而得证2已知数列na的前n项和为nS,若1nnaS(1)求na通项公式;(2)若1111nnncaa,nT为数列 nc的前n项和,求证:21nTn解: (1)由1nnaS,得121a ,则112a ,111(2)nnaSn,两式作差可得:11()0nnnnaaSS,即10nnnaaa,则11(2)2nnana,数列na是首项为12,公比为12的等比数列,则1111( )( )222nnna;证明: (2)111112211nnncaa22212212212224 211412nnnnnnaaaa,22243(1)120nnnaaa ,2212222112(2)43 13nnnnnn
3、aaccaa,又11111222113caa,121223 33nnnc,则122121112(2)(2)(2)(1)3333nnnTnccc112131113313nn ,则21nTn3已知数列na前n项和为nS,232nnSan,*nN,数列 nb是等差数列,11ba,42ba()求数列na, nb的通项公式:()设1,121,2nnnncnab求证:*1211,12ncccnN解: (1)11232aa,12a ,当2n时,232nnSan,11232(1)nnSan,得132nnaa,113(1)nnaa ,113a ,所以1na 是以 3 为首项、3 为公比的等比数列,31nna ;
4、在等差数列 nb中,112ba,428ba,所以2d ,2nbn(2)证明:1,1,21,2,321nnncnn当2n时,由11332 321nnnn得,132130nnn ,所以1113213nnncn,当3n时,212231111( )11111111111193124333242461213nnnccc,当1n 时,1111212c ,当2n 时,12311412cc,所以对于任意的*nN,121112nccc4已知数列 nc的前n项之积为nT,即1 2nnTc cc,且(1)10n nnT,1nnalgc()求数列 nc,na的通项公式;()设数列na的前n项和为nS,12nnnabS
5、,求证:对一切*nN,均有121113nbbb解: ()1 2nnTc cc,2n 时,(1)2(1)1101010n nnnnnnnTcT,又1n 时,11100CT,符合上式,210nnc,121nnalgcn ;()证明:212naaSnnn,312nnnabSn,112211()(11)1111nbn nnnnnnnnnnn ,121111111111()()()132411nbbbnn 11111 123212nn ,得证5已知数列na中,213a ,112nnnnaaa a(1)求数列na的通项公式;(2)令(21)(2)nnan n的前n项和为nT,求证:34nT 解: (1)由
6、213a ,112nnnnaaa a,可得1212112233aaa aa,解得11a ,又对112nnnnaaa a两边取倒数,可得1112nnaa,则1na是首项为 1,公差为 2 的等差数列,可得112(1)21nnna ,所以121nan;(2)证明:由(1)可得(21)11 11()(2)(2)22nnan nn nnn,所以11111111111 323(1.)2324351122 2(1)(2)nnTnnnnnn,因为*nN,所以230(1)(2)nnn,则133224nT 6已知正项数列na的前n项和为nS,且2*2()nnnSaa nN()求数列na的通项公式;()记21na
7、nb ,证明:当*nN时,312122122nnbbbnnbbb()解:由22nnnSaa得21112nnnSaa,则221111222nnnnnnnaSSaaaa,化简得11()(1)0nnnnaaaa,又0na ,故11nnaa当1n 时,解得11a ,因此数列na的通项公式为nan()证明:由题意,21nnb 由于11221nnnbb,且1111110212(21)2nnnn,所以1312211211( ) 1212 (2)(2)(2)2( )21212nnnnbbbbbbbb,化简得312122122nnbbbnnbbb7在等比数列na中,nS为其前n项和,12a ,数列2nnSa是等
8、差数列()求na;()若12aa,证明:*3121122333()(1)(2)(3)()nnnaaaanNS aSaS aSan()解:设公比为q,数列2nnSa是等差数列,12a ,3121322222SSSaaa,22224222qqqqq,解得1q 或2q ,2na 或2nna ()证明:12aa,2nna ,122nnS,121111221232()(22)(2)2(1) 22222(1)2nnnnnnnnnnnnannSannnnn又11232527222nnnnnn,1211221413(1)(2)33aaS aS a 当3n时,312121311223311222527()(1)
9、(2)(3)()(1)(2)22nniiinnaaaaaaiiS aS aS aS anS aS a1411273382nn8已知数列na的前n项和为nS,且1nnaS,*nN,数列 nb满足2lognnba (1)求数列na的通项公式;(2)设221nnnnnabcb b,数列 nc的前n项和为nT,求证:14nT 解: (1)由1nnaS,*nN,可得11111aSaa,解得112a ,2n时,111nnaS,又1nnaS,两式相减可得110nnnnaaSS,即为10nnnaaa,即112nnaa,可得1111( )( )222nnna;数列 nb满足221loglog ( )2nnnban ;(2)证明:222112111(1)222(1) 2nnnnnnnnabncb bn nnn,所以22311 111111.2 22 22 23 22(1) 2nnnTnn11 1112 2(1) 24nn