已知数列an前n项和为sn,sn=n²an-n²(n-1)
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![已知数列an前n项和为sn,sn=n²an-n²(n-1)](/uploads/image/f/4267064-56-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2Csn%3Dn%C2%B2an-n%C2%B2%28n-1%29)
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
a(1)=s(1)=1-5a(1)-85,6a(1)=-84,a(1)=-14.a(n+1)=s(n+1)-s(n)=(n+1)-5a(n+1)-85-[n-5a(n)-85]=1-5a(n+1)+5
S(n+1)=2Sn+3n+1则S(n+1)-Sn=Sn+3n+1即a(n+1)=Sn+3n+1所以Sn=a(n+1)-3n-1所以S(n-1)=an-3(n-1)-1用上式减下式:Sn-S(n-1)
n=1,S1=a1=(a1-1)/3,a1=-1/2;n=2,S2=a1+a2=(a2-1)/3,a2=+1/4;an=Sn-Sn-1=(an-1)/3-(an-1-1)/3=an/3-an-1/32
评析:本页那位热心网友写错了:在得出an+1=3(a(n-1)+1)后,应将a2=8带入求值,因为前面a(n-1),n应大于等于二,所以a1不能算入通项公式中,应检验是否符合n大于等于二时的通项公式,
n=an+1S(n+1)=2Sn+n+5.1Sn=2S(n-1)+n-1+5=2S(n-1)+n+4.2(1)-(2)得S(n+1)-Sn=2[Sn-S(n-1)]+1a(n+1)=2an+1a(n+
当n=1时,a1=-14;当n≥2时,an=Sn-Sn-1=-5an+5an-1+1,所以,又a1-1=-15≠0,所以数列{an-1}是等比数列;(2)由(1)知:,得,从而(nÎN*);
(1)an+2Sn·S(n-1)=0(n≥2),又an=Sn-S(n-1)所以Sn-S(n-1)+2Sn·S(n-1)=0(n≥2)两边同时除以Sn·S(n-1),得1/S(n-1)-1/sn+2=0
1.n=1时,1/S1=1/(1+1)=1/2S1=2n=2时,1/S1+1/S2=1/2+1/S2=2/31/S2=2/3-1/2=1/6S2=6n=1时,S1=2n≥2时,1/S1+1/S2+..
Sn=n-5an-85(1)S(n+1)=n+1-5a(n+1)-85(2)(2)-(1)整理得6a(n+1)=1+5an即a(n+1)-1=(5/6)(an-1)又由S1=a1=1-5a1-85得a
1.n=1时,a1=S1=1²+1=2n≥2时,Sn=n²+nS(n-1)=(n-1)²+(n-1)an=Sn-S(n-1)=n²+n-(n-1)²-
(Ⅰ)由S1=13(a1−1),得a1=13(a1−1)∴a1=−12又S2=13(a2−1),即a1+a2=13(a2−1),得a2=14.(Ⅱ)当n>1时,an=Sn−Sn−1=13(an−1)−
(1)证明:∵Sn=n-5an-85,n∈N*(1)∴Sn+1=(n+1)-5an+1-85(2),由(2)-(1)可得:an+1=1-5(an+1-an),即:an+1-1=56(an-1),从而{
S1=a1=1-1*a12a1=1a1=1/2S2=1-2a2=a1+a2=1/2+a23a2=1/2a2=1/6Sn=1-nanSn-1=1-(n-1)a(n-1)相减an=Sn-Sn-1=1-na
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
(Ⅰ)a1=3,当n≥2时,Sn−1=23an−1+1,∴n≥2时,an=Sn−Sn−1=23an−23an−1,∴n≥2时,anan−1=−2∴数列an是首项为a1=3,公比为q=-2的等比数列,∴
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:
由Sn=13(an−1)可知Sn−1=13(an−1−1),两式相减可得,an=13(an−an−1),即anan−1=−12,(n≥2)故数列数列{an}为等比数列.公比q=−12. 又a
∵数列{an}的通项公式an=2n+1,∴Sn=n(3+2n+1)2=n2+2n,∴Snn=n+2,∴数列{Snn}的前10项的和为10(3+12)2=75.故答案为:75.
Sn-S(n-1)=2An-2A(n-1)=An所以An=2A(n-1)An/2A(n-1)=2即An为等比为2的等比数列令n=1,S1=3+2A1=A1A1=-3所以An=-3*[2^(n-1)]