2.等比数列前n项和性质(1)数列{an}为等比数列,Sn为其前n项和,则Sn,S2nSn,S3-S2,…,仍构成等比数列.(Sn≠0)(2)若某数列前n项和公式为Sn=an-1(a≠0,a≠±1,n∈N),则{an}成等比数列(3)若数列{an}是公比为q的等比数列,则①Sn+m=Sn+qSm②在等比数列中,若项数为2n(n∈N,),则偶S奇 ...
解:设等比数列{an}的公比为q,则Sn,S2n-Sn,S3n-S2n成等比数列,公比为q^n.证明:先证明一个更一般的通项公式.在等比数列中,an=a1q^(n-1)am=a1q^(m-1)两式相除得an/am=q^(n-m),∴an=amq^(n-m).S2n=a1+a2+...+an+a(n+1)+a... 结果...
Sn = a1 * (q^n -1)/(q-1)S2n = a1 * (q^2n -1)/(q-1)S<(k-1)n> = a1 * [q^(k-1)n -1]/(q-1)Skn = a1 * [q^(kn) -1]/(q-1)S<kn> - S<(k-1)n> =[a1/(q-1)]*[q^(kn) - q^(k-1)n]= [a1/(q-1)] * q^[(k-1)n] * (q^n -1...
解答解:当公比q=1时,显然可得Sn,S2n-Sn,S3n-S2n,…构成等比数列; 当q≠1时,Sn=a11−qa11−q(1-qn) S2n-Sn=1−qa11−q(1-q2n-1+qn)=a11−qa11−q(1-qn)qn, 同理可得S3n-S2n=a11−qa11−q(1-q3n-1+q2n)=a11−qa11−q(1-qn)q2n, ...
当公比q=1时,显然可得Sn,S2n-Sn,S3n-S2n,…构成等比数列;当q≠1时,Sn= a1 1-q(1-qn)S2n-Sn= a1 1-q(1-q2n-1+qn)= a1 1-q(1-qn)qn,同理可得S3n-S2n= a1 1-q(1-q3n-1+q2n)= a1 1-q(1-qn)q2n,∴Sn,S2n-Sn,S3n-S2n,…,构成公比为qn的等比数列综上可得Sn,S2n-Sn,S3n-S2n...
解答 解:当公比q=1时,显然可得Sn,S2n-Sn,S3n-S2n,…构成等比数列;当q≠1时,Sn=a11−qa11−q(1-qn)S2n-Sn=a11−qa11−q(1-q2n-1+qn)=a11−qa11−q(1-qn)qn,同理可得S3n-S2n=a11−qa11−q(1-q3n-1+q2n)=a11−qa11−q(1-qn)q2n,∴Sn,S2n-Sn,S3n-S2n,…,构...
1.当q≠1时有Sn=a1(1-q^n)/(1-q),S2n=a1(1-q^2n)/(1-q)S3n=a1(1-q^3n)/(1-q)所以S2n-Sn=a1q^n(1-q^n)/(1-q) S3n-S2n=a1q^2n(1-q^n)/(1-q)所以Sn(S3n-S2n)=[a1q^n(1-q^n)/(1-q)]² (S2n-Sn)²=[a1q^n(1-q^n)/(1-q)]²即Sn(S3n-S2n)=(S2n-Sn...
an = a1.q^(n-1)Sn = a1(q^n -1)/(q-1)[S(2n)- Sn]^2 = [a1(q^(2n) -1)/(q-1) - a1(q^n -1)/(q-1)]^2 =[a1/(q-1)]^2 .(q^(2n)- q^n )^2 =[a1/(q-1)]^2 .[q^(4n)-2q^(3n)+ q^(2n) ]Sn . [S(3n) - S(2n)]=[a1(q^n -1...
数列{an}是首项为m、公比为q(q≠1)的等比数列,Sn是它的前n项和,对任意的n∈N,点(an,S2nSn)( )A. 在直线mx+qy-q=0上B. 在直线qx-my+m=0上C. 在直线qx+my-q=0上D. 不一定在一条直线上
S2n-Sn=a(n+1)+...+a2n=a1*qˆn+a2*qˆn+...+an*qˆn=(a1+...+an)*qˆn=Sn*qˆn S3n-S2n=a(2n+1)+...+a3n=a1*qˆ2n+a2*qˆ2n+...+an*qˆ2n=(a1+...+an)*qˆ2n=Sn*(qˆn)²...