在数列{an}中,a1=5,an+1=3an-4n+2,其中n∈N*.(1)设bn=an-2n,证明数列{bn}是等比数列(2)记数列{an}的前n项和为Sn,若对任意正
解:(1)数列{an}的前n项和为Sn,已知a1=3,Sn+1=3Sn+3①,当n≥2时,Sn=3Sn-1+3②,两式相减,得:an+1=3an,又a1=3,代入Sn+1=3Sn+3,得:a2=9.所以:a_1=8^n.(2)由于b_n=(4n)/(a_(n+1-a_n))=(4n)/(3n+2-3n)=(2n)/(3n),故:T_n=2*1/3+4*1/(3^2)+6*1/...
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. ...
分析(1)通过an=2(n+1)可知数列{an}是首项为4、公差为2的等差数列,进而利用等差数列的求和公式计算可知An=n2+3n,利用3An-Bn=4n可知Bn=3n2+5n,分n=1与n≥2两种情况计算可知bn=6n+2; (2)通过(1)裂项可知cn=1414(1n1n-1n+21n+2),进而并项相加即得结论. ...
(1)证明:∵数列{an}满足a1=1,an+1=3an-4n+2,bn=an-2n,∴bn+1bn=an+1−2(n+1)an−2n=3an−4n+2−2(n+1)an−2n=3,∴数列{bn}是等比数列.(2)∵b1=a1-2=-1.由(1)可得:bn=−1×3n−1=-3n-1.(3)由(1)... (1)由数列{an}满足a1=1,an+1=3an-4n+2,bn=an-2n,可...
解:(1)由an+1=3an-4n+2得an+1-2(n+1)=3(an-2n),又a1-2n=5-2×2=1≠0,an-2n≠0,得(((a_(n+1))-2(n+1)))/(((a_n)-2n))=3,所以,数列{an-2n}是首项为3,公比为3的等比数列,(2)an-2n=3n⇒an=2n+3n,(S_n)=3/2((3^n)-1)+n(n+1),(S_n)-(n^2)-2...
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...
(2)由an+1=3an-4n+2(n∈N*),得an+1-2(n+1)=3(an-2n),从而能证明数列{an-2n}是首项为 1 3,且公比为3的等比数列,由此能求出数列{an}的通项公式.(3)由 1+2bn bn= 1 bn+2= an n= 3n-2+2n n,得 bn= n 3n-2,由此利用错位相减法能求出数列{bn}的前n项和Sn. 解答: 解:(1)...
(2)直接利用拆项法,分别通过等差数列以及等比数列求和求解即可. 解答解:(1)∵a1=1,an+1=3an+4n-2,(n∈N+), ∴an+1+2(n+1)an+2n=3an+4n−2+2(n+1)an+2n=3(an+2n)an+2n=3an+1+2(n+1)an+2n=3an+4n−2+2(n+1)an+2n=3(an+2n)an+2n=3, ...
∵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...