已知数列an的前n项和Sn满足y=r的x次方-1
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an-2/3(-1)^(n-1)=2a(n-1)+4/3(-1)^(n-1)an+2/3(-1)^n=2(a(n-1)+2/3(-1)^(n-1))所以{an+2/3(-1)^n}是等比数列,公比为2
n=1时,S1=a1=2a1-1,a1=1n≥2时,an=Sn-S(n-1)=(2an-1)-(2a(n-1)-1)an=2a(n-1),故an=2^(n-1).
可以用an与Sn之间的关系求当n》2时an=Sn-S(n-1)=2an-2a(n-1)即an=2a(n-1)即数列{an}是等比数列当n=1时a1=S1=2a1-1a1=1an=2的n-1次方
an+2Sn*Sn-1=0其中an=Sn-Sn-1代入上式:Sn-Sn-1+2Sn*Sn-1=0a1=1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得:1/Sn-1-1/Sn+2=0即:1
S1=A1=2A1-3故A1=3而An=Sn-S(n-1)=(2An-3n)-[2A(n-1)-3(n-1)]=2An-2A(n-1)-3故An=2A(n-1)+3故An+3=2[A(n-1)+3]即
Sn=2An-3nS(n-1)=2A(n-1)-3(n-1)两式相减An=2An-3n-(2A(n-1)-3(n-1))An=2A(n-1)-3所以An是等差数列(An-3)/((An-1)-3)=2
由题得:Sn=1-nan于是有:S(n-1)=1-(n-1)a(n-1)两式相减得:an=(n-1)a(n-1)-nan移项后有:(n+1)an=(n-1)a(n-1)于是:an=[(n-1)/(n+
解题思路:其他............................................................解题过程:同学你好,能否把题目写清楚一点
由Sn=n-Sa知,an=Sn-Sn-1=1(>=2).a1=1-Sa
(Ⅰ)证明:由a1+s1=2a1=2得a1=1;由an+Sn=2n得an+1+Sn+1=2(n+1)两式相减得2an+1-an=2,即2an+1-4=an-2,即an+1-2=12(an-2)是首项为
log2(an+1)=n+12^(n+1)=an+1an=2^(n+1)-1
1.证:n=1时,S1=a1=3a1+22a1=-2a1=-1n≥2时,an=Sn-S(n-1)=3an+2-3a(n-1)-22an=3a(n-1)an/a(n-1)=3/2,为定值数列{an}是以
(1)由an+1=Sn+(n+1)①得出n≥2时 an=Sn-1+n②①-②得出an+1-an=an+1整理an+1=2an+1.(n≥2)由在①中令n=1得出a2=a1+2=3,满足a2=
An+2Sn*Sn-1=0Sn-Sn-1+2Sn*Sn-1=01/Sn-1-1/Sn+2=01/Sn=2nSn=1/2n(n>=2)An=1/(2n-2n^2)(n>=2)=1/2(n=1)
因为Sn+Sn-1=3an所以Sn-1+Sn-1+an=3an2Sn-1=2anSn-1=an因为Sn=an+1所以Sn-Sn-1=an+1-anan=an+1-an2an=an+1an+1/an=2
n=1时,a1=S1=a+bn≥2时,Sn=a×n²+bnS(n-1)=a×(n-1)²+b两式相减得:an=Sn-S(n-1)=2a×n-a∴a(n-1)=2a×(n-1)-a∴
解题思路:方法:数列通项的求法:已知sn,求an。求和:错位相减法。解题过程:
an+Sn=2n令n=1a1+S1=2=>a1=1又a(n-1)+S(n-1)=2(n-1)与上式作差an-a(n-1)+an=22an-a(n-1)=2an-2=(1/2)[a(n-1)-2]得证a
an-a(n+1)=ana(n+1)【两边同除以ana(n+1)】得:1/[a(n+1)]-1/[a(n)]=1即:数列{1/(an)}是以1/a1=1为首项、以d=1为公差的等差数列.则:1/[a(