即an+1=-3an,又因为4S1=3a1+4,所以a1=4,故数列{an}是首项为4,公比为-3的等比数列,所以a_n=4•(-3)^(n-1);(2)b_n=(-1)^(n-1)na_n=4n•3^(n-1),所以T_n=4(1•3^0+2•3^1+3•3^2+⋯+n•3^(n-1)),3T_n=4(l•3^1+2•3^2+3•33+⋯+n...
解:(1)数列{an}满足a1=3,an+1=3an﹣4n, 则a2=3a1﹣4=5,a3=3a2﹣4×2=7,…, 猜想{an}的通项公式为an=2n+1. 证明如下:(i)当n=1,2,3时,显然成立, (ii)假设n=k时,ak=2k+1(k∈N+)成立, 当n=k+1时,ak+1=3ak﹣4k=3(2k+1)﹣4k=2k+3=2(k+1)+1,故n=k+1时成立, 由...
(1)Sn=3(1-an)得Sn-1=3-3an-1(n≥2)则Sn-Sn-1=an=-3an+3an-1∴an=34an−1当n=1时,S1=3-3a1=a1∴a1=34∴{an}为等比数列,且a1=34,q=34∴an=34•(34)n−1=(34)n…(5分)(2)由bn=4n-1-3bn-1(n≥2)得b... (1)由Sn=3(1-an)得Sn-1=3-3an-1(n≥2),利用递推公...
①-②:an+1-an=3(Sn-Sn-1)=3an,∴an+1=4an,n>1…(4分)∵a2=3S1+1=3a1+1=3t+1,a1=t,∴3t+1=4t,∴t=1∴当t=1时,a2=4a1,数列{an}是等比数列…(6分)(Ⅱ)在(Ⅰ)的结论下,an+1=4an,∴an+1=4n,…(8分)∴bn=log4an+1=n,…(9分)cn=an+bn=4n−1+n,…(10分)...
解答(1)解:∵2Sn=3an-4n+3,∴2Sn+1=3an+1-4(n+1)+3,两式作差得:an+1=3an+4;(2)证明:由an+1=3an+4,得an+1+2=3(an+2),即bn+1=3bn,由2Sn=3an-4n+3,得2a1=3a1-4+3,即a1=1.又b1=a1+2=3≠0.∴{bn}是以3为首项,3为公比的等比数列. 点评 本题考查数列递推式,考查了等比关...
整理得an=3an-1+4,转换为an+2=3(an-1+2),故数列中的λ=2,数列{an+λ}为等比数列,首项为6,公比为3.(2)由(1)得:a_n+2=6*3^(n-1)=2*3^n,所以b_n=n•3^n,故T_n=1*3+2*3^2+3*3^3+...+n•3^n,①,3T_n=1*3^2+2*3^3+...+n•3^(n+1),②,②-①得...
(17)答案: |a_1=-3⋅(3/4) (Ⅱ)-3.1解:(1)当n=1时,4(a1+a2)=3a1-9.已知a1= -9/4 f'(1)+1,1,2,-9/4-9=-(27)/4 9 解得 u_1=-(27)/(16) 1分 16 当n≥2时.由4S =3S -91.得4S =3S -92. 1-2得4an=3an.(3分) 27 3 因为 u_2=-(27)/(16)...
(1)∵3Sn=4an-4 ∴n=1时,3a1=4a1-4,∴a1=4.当n≥2时,3Sn-1=4an-1-4,∴3an=3Sn-3Sn-1=4an-4an-1,即anan-1=4,∴{an} 是一个首项为4,公比为4的等比数列,∴an=4n,∴bn=log2an=log24n=2n.(2)cn=b1+b2+…... 解析看不懂?免费查看同类题视频解析查看解答 特别推荐 热点考点 ...
求通项公式1.数列{an}的前几项和Sn=n方-3n,则通项公式an为()A.2n-4 B.2n-3 C.4n-5 D.4n-32.等差数列{an}k ,若a1005+a1007+a1009=6,则该数列前2013项的和为多少3.已知数列{an}的首项a1=2,an+1=3an+2,数列{a
∵n=1时也成立,∴an=4n-3an+1-an=[4(n+1)-3]-[4n-3]=4∴{an}成等差数列 (2)a1=S1=2a2=S2-S1=5a3=S3-S2=9∵a2-a1≠a3-a2∴{an}不是等差数列. 【点评】已知Sn,求an,要注意a1=S1,当n≥2时an=Sn-Sn-1, 因此an= . 练习:已知等差数列{a }的前项和S 满足条件:S =2n +3n, 求此...