前n项和为sn,且2根号下sn=an 1
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1.Sn=-2an+3有S(n-1)=-2a(n-1)+3则an=Sn-S(n-1)=-2an+2a(n-1)=>an=a(n-1)*2/3所以,{an}为共比数列,q=2/32.Sn=-2an+3有
(1)证明:数列{根号下Sn}是一个等差数列:(2)求{an}通项公式证明:(1)当n=1时,S1=a1=1,√S1=1当n≥2时,an=(√Sn+√Sn-1)/2=Sn-Sn-1(√Sn+√Sn-1
Sn+1/(2n+1)-Sn/(2n-1)=1Sn/(2n-1)=S1+n-1→Sn=(S1+n-1)(2n-1)→Sn=n(2n-1)an=4n-31/√an=2/2√(4n-3)>2/(√4n-3
1.n≥2时,an=Sn-S(n-1)=√Sn+√S(n-1)[√Sn+√S(n-1)][√Sn-√S(n-1)]=√Sn+√S(n-1)[√Sn+√S(n-1)][√Sn-√S(n-1)-1]=0算
设:等差数列{an}的公差为d,通项为an=a1+(n-1)d,则:sn=a1+a2+...+an=na1+n(n-1)d/2lim(n->∞)(n*an)/Sn=lim(n->∞)[n*(a1+(n
由题意得,Sn=[(an+1)/2]^2①则S(n+1)=[(a(n+1)+1)/2]^2②②-①得(结合a(n+1)=S(n+1)-Sn)a(n+1)=[(a(n+1)+1)/2]^2-[(an+1
an,根号下2Sn,a(n+1)成等比数列,即2Sn=an*a(n+1)令n=12S1=2a1=a1*a2,得a2=2当n>=2时,将2Sn=an*a(n+1),改写为2S(n-1)=a(n-1)*a
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
由题意得,Sn=[(an+1)/2]^2①则S(n+1)=[(a(n+1)+1)/2]^2②②-①得(结合a(n+1)=S(n+1)-Sn)a(n+1)=[(a(n+1)+1)/2]^2-[(an+1
当n=1时2s1=2a1=a1+1/a1a1=1/a1a1²=1{an}是正整数数列a1=1=(根号下1)-(根号下0)满足如果a(k)=(根号下k)-(根号下k-1)2S(k)=a(k)+
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
题目应有笔误,应该是“设数列{an}各项为正数,前n项和为Sn,且二倍根号下Sn=an+1,(n为一切正整数)(1)求数列{an}通项公式(2)记bn=1/(二倍根号下an+二倍根号下an+1),求数
1.2√Sn=an+14Sn=(an)^2+2an+14S1=(a1)^2+2a1+1=4a1,a1=14S(n-1)=[a(n-1)]^2+2a(n-1)+14an=4[sn-s(n-1)]=(an
2√Sn=an+1则有,4Sn=(an+1)²4a(n+1)=4[S(n+1)-Sn]=[a(n+1)+1]²-(an+1)²=[a(n+1)]²+2a(n+1
因为2√S(n)=a(n)+12√S(n+1)=a(n+1)+1所以两式平方相减4(S(n+1)-S(n))=[a(n+1)+1]^2-[a(n)+1]^24·a(n+1)=[a(n+1)]^2+2·
2根号Sn=an+14Sn=an的平方+2an+14Sn_1=an_1的平方+2an_1+1〔n≥2〕又Sn-Sn_1=an所以4an=an的平方+2an-an_1的平方-2an_1划简为〔an+an
1.a[n]=S[n]-S[n-1]=1/2(√S[n]+√S[n-1])==>√S[n]-√S[n-1]=1/2==>√S[10]-√S[4]=1/2*6=3,√S[4]=√4=2==>√S[10]
2*Sn^(1/2)=An+1(1)2*S1^(1/2)=A1+1,S1=A1A1=1(2)Sn=(An+1)^2/4S(n-1)=[A(n-1)+1]^2/4An=Sn-S(n-1)=(1/4)*(
根号下Sn-根号下S(n-1)-根号2=0根号下Sn-根号下S(n-1)=根号2设bn=(Sn)^(1/2)则:bn-b(n-1)=根号2b1=(S1)^(1/2)=(a1)^(1/2)=根号2bn=
解题思路:考查数列的通项,考查等差数列的证明,考查数列的求和,考查存在性问题的探究,考查分离参数法的运用解题过程: