已知数列bn的前n项和Sn=3 2n²-1 2n
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A(n+1)=S(n+1)-Sn得:S(n+1)-Sn=Sn+3^n∴S(n+1)=2Sn+3^n∴S(n+1)-3*3^n=2Sn-2*3^n∴S(n+1)-3^(n+1)=2(Sn-3^n)∴B(
由an+1=Sn+3n得:Sn+1-Sn=Sn+3n,即Sn+1=2Sn+3n.所以Sn+1-3n+1=2Sn+3n-3n+1.整理得:Sn+1-3n+1=2(Sn-3n),这就是说,数列{Sn-3n
Sn=2n^2+2n=>Sn+1=2(n+1)^2+2(n+1)=>an=4n+4T1=2-b1=>b1=1b1+b2+b3+.+bn=2-bn=>Tn-1=2-2bn=>bn=1/2^(n-1)Tn
n=1,S1=a1=2,n>1,an=Sn-S(n-1)=2n,n=1时也适合,故:an=2nbn=(1/4)·1/n(n+1)4bn=1/n(n+1)=1/n-1/(n+1),所以:4Tn=[(1-
第一问,前一项减后一项(简单的)an=4n,bn=1/2^(n-1)所以cn=(4n)^2*1/2^(n-1)因cn均大于0,所以c(n+1)/cn=(n+1)^2/2n^2因(n+1)^2-2n^2
(1)a1=S1=3-1=2n>1时,an=Sn-S(n-1)=3*(3/2)^(n-2)*(3/2-1)=(3/2)^(n-1)n=1不符合此式,故an=2,n=1an=(3/2)^(n-1),n>
n=1时,S1=a1=2-1=1n≥2时,Sn=2n-n^2S(n-1)=2(n-1)-(n-1)^2an=Sn-S(n-1)=2n-n^2-2(n-1)+(n-1)^2=3-2nn=1时,a1=3-
Sn=2n^2+2n当n=1时an=a1=S1=2+2=4当n≥2且n∈N*时an=Sn-S(n-1)=2n^2+2n-2(n-1)^2-2(n-1)=2n^2+2n-2(n^2-2n+1)-2n+2
(1)n=1时,S1=1-a1所以a1=1/2an=Sn-S(n-1)=1-an-(1-a(n-1)=a(n-1)-an所以:an=1/2a(n-1),{an}是等比数列an=(1/2)^n(2)Tn
因为Sn=2n-n^2所以S(n-1)=2(n-1)-(n-1)^2两式相减an=3-2n所以bn=5^(3-2n)=5*(1/25)^(n-1)所以bn是以5为首项1/25为公比的等比数列前n项和=
(1)如果an=n,bn=(1/3)*n,则an/bn=3,因此Sn=3n;(2)如果an=n,bn=1/(3n),那么an/bn=3n^2,因此Sn=n(n+1)(2n+1)/2.(有公式1^2+2
An是等差数列,通项An=6n+2Bn是等比数列,通项Bn=1/8^(2n-3)An+logxBn=6n+2+logx8^(-2n+3)=6n+2-(2n-3)logx8要想为常数,上式得与n无关,所
Sn=n^2推出an=2n-1bn=(2n-1)/3^nTn=b1+b2+b3+……+bn-1+bn=1/3+3/3^2+5/3^3+……+(2n-3)/3^n-1+(2n-1)/3^n①3Tn=1+
根据A1=S1(n=1);An=Sn-Sn-1(n>=2)可得An=2n-1;进而得Bn=(2n-1)/3^n下证Tn=1-(n+1)/3^n显然T1=1/3=B1Tn-Tn-1=1-(n+1)/3^
由Sn=2n-n^2可得Sn-1=2(n-1)-(n-1)^2Sn-Sn-1=an=3-2nbn=5^(3-2n)=5*(1/25)^(n-1)所以{bn}是以5为首项1/25为公比的等比数列数列{b
再答:满意采纳,不懂追问,谢谢
n=n(n+1)=n^2+nSn=b1+b2+...+bn=(1^2+1)+(2^2+2)+...+(n^2+n)=(1^2+2^2+...+n^2)+(1+2+...+n)=n(n+1)(2n+1)
Sn+1=an+1+Sn.又an+1=Sn+3∧n.Sn+1=2Sn+3∧n.①3∧n+1=3×3∧n.②①-②得Sn+1-3∧n+1=2(Sn-3∧n)后面的你知道吧,我就不说了.
n=b1.q^(n-1)bn=an-3nan=bn+3n=b1.q^(n-1)+3nSn=a1+a2+...+an=b1(q^n-1)/(q-1)+3n(n+1)/2
Sn=n^2S(n-1)=(n-1)^2an=Sn-S(n-1)=n^2-(n-1)^2=2n-1因此得到数列{an}的通项公式为an=2n-1