数列an的前n项和Sn满足log以2为底Sn-1的对数=n 1则an=
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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
2·a(n)=2[Sn-S(n-1)]=(n+1)an-n·a(n-1)∴(n-1)an=n·a(n-1),∴an/[a(n-1)]=n/(n-1),.,a3/a2=3/2,a2/a1=2/1,将上述
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
(1)Sn=n^2-10nan=Sn-S(n-1)=(2n-1)-10=2n-11=>{an}是等差娄列(2)bn=(an+1)/an=(2n-10)/(2n-11)maxbn=b1=8/9minbn
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+
我就说第二问吧.若{an}中存在三项,它们可以构成等差数列,则有2an=(an-1)+(an+1)即2*(3*2^n-3)=3*2^(n+1)-3+3*2^(n-1)-3,3*2^(n+1)-6=3*
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
若存在某一Sk=0,必有ak=0,从而S(k-1)=Sk-ak=0同理推出a(k-1)=a(k-2)=……=a1=0a1=0与已知a1=1矛盾所以不存在Sk=0,Sn恒不为零由An=2(Sn^2)/(
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∴
Sn=n^2-8n+1sn-1=n^2-10n+10相减得an=2n-9当n《4时,bn=9-2n当n>4时,bn=2n-9
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
(1)当n=1时,T1=2S1-1因为T1=S1=a1,所以a1=2a1-1,求得a1=1(2)当n≥2时,Sn=Tn-Tn-1=2Sn-n2-[2Sn-1-(n-1)2]=2Sn-2Sn-1-2n+
答案:(n^-2n+3)*2^(n+1)-6证明可用数学归纳法