Sn=1 2(an 1 an)
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当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有an+1=3an,为等比数列,且公比为3,所
∵1+12+14+…+(12)n-1=1−(12)n1−12=2−12n−1,∴Sn=2n−(1+12+122+…+12n−1)=2n-1−12n1−12=2n-2+12n−1.
Sn-S(n-m)=A(n-m+1)+A(n-m+2)+……+A(n-m+m)=b共m项A(n-m+1)=A1+(n-m)dA(n-m+2)=A2+(n-m)d……A(n-m+m)=An=Am+(n-
当公比为1时,Sn=n,数列{Sn+12}为数列{n+12}为公差为1的等差数列,不满足题意;当公比不为1时,Sn=1−qn1−q,∴Sn+12=1−qn1−q+12,Sn+1+12=1−qn+11−
由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn
S3=a1+a2+a3=3a2=12a2=4设公差为d,则a1=a2-d=4-da3=a2+d=4+d2a1、a2、a3+1成等比数列,则a2²=(2a1)(a3+1)2(4-d)(4+d+
(1)∵数列{an}的前n项和为Sn,且an=12(3n+Sn)对一切正整数n成立∴Sn=2an-3n,Sn+1=2an+1-3(n+1),两式相减得:an+1=2an+3,∴an+1+3=2(an+
Sn=12n-n^2Snmax=36Sn=12n-n^2Sn-1=12(n-1)-(n-1)^2两式相减an=12-2n+1=-2n+13数列{|An|}的前n项和Tn当n6时Tn=36+1+3+5+
{an}是等差数列S3=a1+a2+a3=3a2=12a2=4设公差为da1=4-da3=4+d2a1,a2,a3+1成等比数列(a2)^2=2a1·(a3+1)4^2=2(4-d)(4+d+1)8=
(Ⅰ)证明:把n=1代入Sn=2an+3n-12,得a1=2a1+3-12,解得a1=9,当n≥2时,an=Sn-Sn-1=(2an+3n-12)-[2an-1+3(n-1)-12]=2an-2an-
An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(
由题意,S(n)-S(n-1)=2a(n+1)-2a(n),即a(n)=2a(n+1)-2a(n),于是a(n+1)=a(n)*3/2,即a(n)是公比是q=3/2的等比数列,且首项是a(1)=1,所
(1)n≥2,sn2=(sn-sn-1)(sn-12)∴sn=sn−12sn−1+1即1sn-1sn−1=2(n≥2)∴1sn=2n-1故sn=12n−1(2)bn=sn2n+1=1(2n+1)(2n
再问: 再问:那个划横线的答案是不是错了再答:我觉得是
(1)由sn=sn-12sn-1+1(n≥2),a1=2,两边取倒数得1Sn=1Sn-1+2,即1Sn-1Sn-1=2.∴{1sn}是首项为1S1=1a1=12,2为公差的等差数列;(2)由(1)可得
设等差数列{an}的公差为d,由题意得a22=2a1(a3+1)3a1+3×22d=12,解得a1=1d=3或a1=8d=−4,∴sn=12n(3n-1)或sn=2n(5-n).
(1)Sn/n=-n+12=>Sn=-n²+12n(2)an=Sn-S(n-1)=-n²+12n+(n-1)²-12(n-1)=-2n+1+12=-2n+13所以an-a
(1)x=ny=Sn代入函数方程Sn=-n²+12n为数列的项数,n为正整数,所求函数表达式为:Sn=-n²+12(n∈N+)/注意:表达式不要忘了n的定义域(2)n=1时,a1=
(1)An=3(1+2^n)(2)由题知,Sn=2An+3n-12=6(2^n-1)+3nBn=(An-3)/(Sn-3n)(A(n+1)-6)=(3*2^n)/(6(2^n-1))(3(2^(n+1