已知数列{an}的前n项和为Sn,且满足Sn=3 2(an-1)(n属于正整数)
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![已知数列{an}的前n项和为Sn,且满足Sn=3 2(an-1)(n属于正整数)](/uploads/image/f/4267615-31-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3Sn%3D3+2%28an-1%29%28n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%29)
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)
因为Sn-Sn-1=n^2-3n-{(n-1)^2-3(n-1)}=2n-4.又由an=Sn-Sn-1,所以an=2n-4,最后还要验证一下,当n=1时,S1=a1,符合题意.d=an-an-1=2易
a1=S1=3+2=5,an=Sn-Sn-1=(3+2n)-(3+2n-1)=2n-1,当n=1时,2n-1=1≠a1,∴an=5,n=12n−1,n≥2.
S(n-1)=2a(n-1)-1所以Sn-S(n-1)=2an-2a(n-1)因为Sn-S(n-1)=an所以an=2an-2a(n-1)所以an=2a(n-1)an/[a(n-1]=2所以an是等比
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+
(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+..
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
1:已知数列{an}的前n项和是S=32n-n(平方),求数列{|an|}的前n项和Tn.因为.an=sn-sn-1,S=32n-n^2=32n-n^2-32n+32+n^2-2n+1
a(n+1)=s(n)+3(n+1);an=s(n-1)+3n;两边同减a(n+1)-an=s(n)-s(n-1)+3=an+3所以a(n+1)=2*an+3bn=an+3an=bn-3a(n+1)=
因为数列a1,a2-a1,a3-a2,a4-a3.是首相为1公比为2的等比数列则an所以a1,a2-a1,a3-a2,a4-a3.an-a(n-1)的前项和为a1+a2-a1+a3-a2+a4-a3+
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=kq^n-k∴S(n+1)=kq^(n+1)-k∴a(n+1)=S(n+1)-Sn=[kq^(n+1)-k]-(kq^n-k)=k[q^(n+1)-q^n]=k[(q-1)q^na(n+1)/
Sn=1/3(an-1)Sn-1=1/3(an-1-1)Sn-Sn-1=1/3(an-an-1)即an=1/3(an-an-1)然后应该会了吧,可惜我用电脑不如手写的灵活,看看会了吗
因为(n,Snn)在y=3x-2的图象上,所以将(n,Snn)代入到函数y=3x-2中得到:Snn=3n−2,即{S}_{n}=n(3n-2),则an=Sn-Sn-1=n(3n-2)-(n-1)[3(
当n=1时,a1=S1=1当n≥2时,an=Sn-S(n-1)=3n²-2n-3(n-1)²+2(n-1)=6n-5∵当n=1时,满足an=6n-5又∵an-a(n-1)=6n-5
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)]