数列求和专题

1.数列{an}的前n项和为Sn,若)1(1nnan,则S5等于()A.1B.65C.61D.3012.等差数列{an}的前n项和记为Sn,若a2+a6+a10为一个确定的常数,则下列各数中也是常数的是()A.S6B.S11C.S12D.S133.如果f(a+b)＝f(a)·f(b)且f(1)＝2,则)2007()2008()5()6()3()4()1()2(ffffffff等于()A.4016B.1004C.2008D.20064.数列841,631,421,2112222,…的前n项和等于________.5.已知数列{log2(an-1)}为等差数列,且a1＝3,a2＝5.(1)求证:数列{an-1}是等比数列;(2)求nnaaaaaa12312111的值.6.设{an}是等差数列，证明以naaabnn21(n∈N*)为通项公式的数列{bn}是等差数列.7.已知正项数列{an}满足an+12-an2-2an+1-2an＝0,a1＝1.设bn＝n3-3n2+5-an.求数列{an},{bn}的通项公式;8.已知数列{an}的前n项和为Sn，a1＝1,数列{an+Sn}是公差为2的等差数列.(1)求a2,a3;(2)证明数列{an-2}为等比数列;(3)求数列{nan}的前n项和Tn.9.设数列{an}的前n项和为Sn,已知ban-2n＝(b-1)Sn.(1)证明当b＝2时,{an-n·2n-1}是等比数列;(2)求{an}的通项公式.4.)2)(1(23243nnn5.(1)证明：设log2(an-1)-log2(an-1-1)＝d(n≥2),∴d＝log2(a2-1)-log2(a1-1)＝log24-log22＝1.∴log2(an-1)＝n.∴an-1＝2n.∴2111nnaa(n≥2).∴{an-1}是以2为首项,2为公比的等比数列.(2)解:由(1)可得an-1＝(a1-1)·2n-1,∴an＝2n+1.∴nnnnaaaaaa221221221111123212312nn2112121212.6.设等差数列{an}的公差是d(常数),∴1121211naaanaaabbnnnn)1(2))(1(2)(111naannaannn)(21221111nnnnaaaaaad21(常数),其中n≥2.∴{bn}是等差数列.7.解:(1)由an+12-an2-2an+1-2an＝0,得(an+1+an)(an+1-an-2)＝0.因为an＞0,所以an+1-an-2＝0,an+1-an＝2.所以数列{an}是以a1＝1为首项,以2为公差的等差数列.所以an＝1+(n-1)×2＝2n-1,bn＝n3-3n2+5-an＝n3-3n2+5-2n+1＝n3-3n2-2n+6.8.(1)解:∵数列{an+Sn}是公差为2的等差数列,∴(an+1+Sn+1)-(an+Sn)＝2,即221nnaa.∵a1＝1,∴47,2332aa.(2)证明:由题意,得a1-2＝-1,∵212222221nnnnaaaa,∴{an-2}是首项为-1,公比为21的等比数列.(3)解:由(2)得an-2＝-(21)n-1,∴nan＝2n-n·(21)n-1.∴Tn＝(2-1)+(4-2·21)+［6-3·(21)2］+…+［2n-n·(21)n-1］.∴Tn＝(2+4+6+…+2n)-［1+2·21+3·(21)2+…+n·(21)n-1］.设12)21()21(32121nnnA,①∴nnnA)21()21(3)21(2212132.②由①-②,得nnnnA)21()21()21(2112112,∴nnnnA)21()21(1)21(121.∴An＝4-(n+2)·(21)n-1.∴4)1()21()2(4)21()2(2)22(11nnnnnnTnnn9..解:由题意，知a1＝2,且ban-2n＝(b-1)Sn,ban+1-2n+1＝(b-1)Sn+1,两式相减,得b(an+1-an)-2n＝(b-1)an+1,即an+1＝ban+2n.①(1)证明:当b＝2时,由①，知an+1＝2an+2n.于是an+1-(n+1)·2n＝2an+2n-(n+1)·2n＝2(an-n·2n-1).又a1-1×21-1＝1≠0,所以{an-n·2n-1}是首项为1,公比为2的等比数列.(2)当b＝2时,由(1),知an-n·2n-1＝2n-1,即an＝(n+1)2n-1.当b≠2时,由①得)221(222212221111nnnnnnnnnbabbbbabbaba,因此nnnnbbbbabba2)1(2)221(2211111,得.2],)22(2[21,1,21nbbbnannn