An的Sn=n^2 2n,Bn=2^n×An则Bn的前n项和

S_n=n^2+n,S_(n-1)=〖(n-1)〗^2+n-1,∴a_n=S_n-S_(n-1)=2n (n>1),验证当n=1时,a_1=S_1=2,∴n=1时亦立,∴a_n=2n,

Sn=-an-(1/2)^(n-1)+2所以S(n-1)=-a(n-1)-(1/2)^(n-2)+2相减Sn-S(n-1)=an=-an-(1/2)^(n-1)+a(n-1)+(1/2)^(n-2)(

an/bn={[a1+a(2n-1))]/2}/{[b1+b(2n-1)]/2}=n{[a1+a(2n-1))]/2}/n{[b1+b(2n-1)]/2}=S(2n-1)/T(2n-1)=2(2n-1

∵{an}与{bn}是等差数列∴Sn=[n(a1+an)]/2Tn=[n(b1+bn)]/2∴Sn/Tn=(a1+an)/(b1+bn)∵等差数列{an}与{bn}的前n项和的比为2n:(3n+1)∴

(1)Sn=n^2-10nan=Sn-S(n-1)=(2n-1)-10=2n-11=>{an}是等差娄列(2)bn=(an+1)/an=(2n-10)/(2n-11)maxbn=b1=8/9minbn

S(2n-1)=(A1+A(2n-1))×(2n-1)/2=(A1+A1+(2n-2)d)×(2n-1)/2=(A1+(n-1)d)×(2n-1)=An×(2n-1)同理T(2n-1)=Bn×(2n-

（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

哦,TN=(+1)^/2

本题考查的是数列的性质a1+a2n-1=2an因为S2n-1=[(n+1)(a1+a2n-1)]/2=(n+1)anT2n-1=[(n+1)(b1+b2n-1)]/2=(n+1)bn所以an/bn=S

{an}是等差数列,a2=a1+da3=a1+2d....an=a1+(n-1)da(2n-1)=a1+(2n-2)da1+a(2n-1)=2a1+(2n-2)d2an=2a1+2(n-1)d=2a1

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

设Sn=k(7n^2+n)an=Sn-S(n-1)=k(14n-6)Tn=k(4n^2+27n)bn=Tn-T(n-1)=k(8n+23)an:bn==(14n-6)/(8n+23)再问：错·再答：哪

Sn/Tn=2n/(3n+1)(a1+a1+(n-1)*d1)/(b1+b1+(n-1)*d2)=2n/(3n+1)(2a1-d+n*d1)/(2b1-d2+n*d2)=2n/(3n+1)->2a1=

再答：满意采纳，不懂追问，谢谢

Sn=((An+1)/2)^2A1=S1=((A1+1)/2)^2(A1-1)^2=0A1=1Sn=n(A1+An)/2=n(1+An)/2=((An+1)/2)^2(An+1)/2=nAn=2n-1

因为这里的Sn和Tn只知道一个比值,而不是Sn就等于2n,Tn就等于3n+1,所以如果要用an=sn-s(n-1),那么必须求出Sn【事实上这里的Sn=2n（假定）,Tn都差了n倍或者2n,3n...

∵{an}与{bn}是等差数列∴Sn=[n(a1+an)]/2Tn=[n(b1+bn)]/2∴Sn/Tn=(a1+an)/(b1+bn)∵等差数列{an}与{bn}的前n项和的比为2n:(3n+1)∴

Sn=n^2-8n+1sn-1=n^2-10n+10相减得an=2n-9当n《4时,bn=9-2n当n>4时,bn=2n-9

等差数列数列的性质a1+a[2n-1]=2an因为S[2n-1]=[(2n-1)(a1+a[2n-1])]/2=(2n-1)anT[2n-1]=[(2n-1)(b1+b[2n-1])]/2=(2n-1

a1=S1=1²=1Sn=n²Sn-1=(n-1)²an=Sn-Sn-1=n²-(n-1)²=2n-1n=1时,a1=2-1=1,同样满足.数列{an