正项数列{an}满足2Sn=an^2 n-4
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10Sn=(an)²+5an+610S(n-1)=(a(n-1))²+5a(n-1)+6两式相减,得5a(n-1)+5an=(an)²-(a(n-1))²5=a
(1)n=1时,a1=S1=1/2*a1^2+1/2*a1,解得a1=1,当n>=2时,an=Sn-S(n-1)=1/2*an^2+1/2*an-1/2*[a(n-1)]^2-1/2*a(n-1),化
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
数列为正项数列,则Sn>0n≥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-
解:(1)S1=a1=1;(先求出前4项再猜)S2=a1+a2=2^2×a2=4a2;a2=(1/3)a1=1/3;S2=a1+a2=4/3S3=a1+a2+a3=3^2×a3=9a3;a1+a2=8
S[1]=a[1]=1/2(a[1]+1/a[1]),于是:a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是:a[2]=√2-1,S[2]=√2S[3]=a[3]
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
a(1)=1a(2)=√2-1a(3)=√3-√2a(4)=2-√3猜想a(n)=√n-√(n-1)
根据2Sn=an^2+n得到2a1=a1^2+1求得a1=1或a1=-1又因为an>0所以a1=1同理求得a2=2a3=3(2)猜想an=n证明:因为2Sn=an^2+n……①那么2Sn-1=an-1
2Sn=a(n)²+a(n)所以2S(1)=a(1)²+a(1)即2a(1)=a(1)²+a(1)因为a(1)=S(1)>0,a(1)=12Sn=a(n)²+a
Sn=(an+1)^2/4=(an^2+2an+1)/4Sn-1=[a(n-1)+1]^2=[(a(n-1)^2+2a(n-1)+1]/4Sn-Sn-1=an=[an^2+2an-a(n-1)^2-2
an=5n-310Sn=an^2+5an+610S(n+1)=a(n+1)^2+5a(n+1)+6两式相减得a(n+1)^2-an^2=5a(n+1)+5an左右同除a(n+1)+an得a(n+1)-
a2n+an-2Sn=0(1)a2(n-1)+a(n-1)-2S(n-1)=0(n≥2)(2)(1)-(2),得a2n+an-2Sn-a2(n-1)-a(n-1)+2S(n-1)=a2n-a2(n-1
2a[n]-n-1=a[n-1]【1】待定系数:2(a[n]+xn+y)=a[n-1]+x(n-1)+y【2】将【1】式a[n-1]代入上式:(注意:也可变换后用a[n]代入上式,看方便确定)2(a[
由a1=S1=1/6(a1+1)(a1+2),解得a1=1或a1=2,由假设a1=S1>1,因此a1=2,又由a(n+1)=S(n+1)-Sn=1/6(a(n+1)+1)(a(n+1)+2)-1/6(
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
an=Sn-Sn-1(n>=2)an=1/2a(n-1)-1/2a(n-2)=(1/2)a将a=1代入an不符,则该数列以分段的形式构成an=1(当n=1),an=1/2a(n>=2)
题目条件应为:Sn=3an+2an=Sn-S(n-1)(n≥2)=3an-3a(n-1)(n≥2)=>an/a(n-1)=3/2.∴数列{an}成等比数列当n=1时,a1=3a1+2a1=-1.=>a
(An)^2=2Sn-An=>(A(n-1))^2=2S(n-1)-A(n-1)=>(An)^2-(A(n-1))^2=2Sn-An-2S(n-1)+A(n-1)=>(An+A(n-1))*(An-A