如何证明Sn,S2n-Sn,S3n-S2n为等比
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![如何证明Sn,S2n-Sn,S3n-S2n为等比](/uploads/image/f/3530076-60-6.jpg?t=%E5%A6%82%E4%BD%95%E8%AF%81%E6%98%8ESn%2CS2n-Sn%2CS3n-S2n%E4%B8%BA%E7%AD%89%E6%AF%94)
因为数列{an}为等差数列,且a1=1,则由等差数列性质可得:前n项和Sn=a1n-(n(n-1)/2)*D即Sn=n-(n(n-1)/2)*D,S2n=2n-(2n(2n-1)/2)*D且S2n/S
an=a1q^(n-1)Sn=a1(q^n-1)/(q-1)(Sn)^2+(S(2n))^2=[a1(q^n-1)/(q-1)]^2+[a1(q^(2n)-1)/(q-1)]^2=[a1/(q-1)]
(1)Sn=48=a1+a2+……+an=48S2n=a1+a2+……an+a(n+1)+……a(2n)=60Sn+Sn*q^n=S2nq^n=1/4S3n=a1+……a(3n)=S2n+a(2n+1
①依题意知sns2n-sns3n-s2n成等比数列设s3n=x那么有48*(x-60)=(60-x)^2解出x即可②在原式两边同时减去2得an-1=2a(n-1)-2即an-1/[a(n-1)-1]=
Sn,S2n-Sn,S3n-S2n成等比数列48,12,3S3n-S2n=3S3n=3+S2n=63
法一:Sn,S2n-Sn,S3n-S2n成等比数列∴(S2n-Sn)/Sn=(S3n-S2n)/(S2n-Sn)∴S3n=63法二:因为本题与n无关,不妨设n=1,则有a1=s1=36,a2=s2-s
证明:∵已知等比数列的前n项,前2n项,前3n项∴S[n]=a[1](1-q^n)/(1-q)S[2n]=a[1][1-q^(2n)]/(1-q)S[3n]=a[1][1-q^(3n)]/(1-q)∵
根据等差数列的性质:Sn,S[2n]-S[n],S[3n]-S[2n]成等差数列由已知:Sn=10S[2n]-S[n]=20所以:S[3n]-S[2n]=30又S[2n]=30所以S[3n]=60再问
Sn=a1(1-q^n)/1-q=48s2n=a1(1-q^2n)/1-q=60那么s2n/sn=1+q^n=60/48q^n=1/4q^2n=1/16q^3n=1/64s3n=a1(1-q^3n)/
等比数列的Sn,S2n-Sn,S3n-S2n也成等比数列.即48,12,S3n-60成等比S3n-60=12^2/48=3s3n=63
特殊:令n=1,a1=30,a2=70,则a3=110,s3n=210一般:s2n-sn=70,楼上说的应是这个30,70,……下一个是110s3n=s2n+(s3n-s2n)=100+110=210
Sn,S2n,S3n,成等差数列,2*S2n=Sn+S3nS3n=72补充,像S1,Sk,S2k,S3k,下标成等差数列,则这三个数成等差数列再问:已知等差数列{an}中,a3=-2,a8=8,求a1
设等差数列{an}的首项为公差为dSn=a1+a2+……+anS2n-Sn=an+1+an+2+……+a2nS3n-S2n=a2n+1+a2n+2+……+a3n(S2n-Sn)-Sn=(an-a1)+
S2n=1+1/2+...+1/n+(1/n+1)+(1/n+2)+.+1/2nSn=1+1/2+...+1/n所以:S2n-Sn=(1/n+1)+(1/n+2)+.+1/2n
我擦,这个缩写我竟然看不懂~·
等差数列中,Sn、S2n-Sn、S3n-S2n...的公差为n^2*d等比数列中,Sn、S2n-Sn、S3n-S2n...的公比为q^n
当n≥2时,可以化为Sn-S(n-1)=-2Sn×S(n-1),两边同除以Sn×S(n-1),得1/Sn-1/S(n-1)=2所以{1/Sn}是以2为首项,2为公差的等差数列即1/Sn=2nSn=1/
S2n=2n+n*(2n-1)dSn=n+n(n-1)d/24Sn=4n+2(n^2-n)dS2n/Sn=4S2n=4Sn4n+2d(n^2-n)=2n+(2n^2-n)d整理,得dn=2nd=2S2
(1)S(2n)=2n(a1+a2n)/2=n(a1+a2n)Sn=n(a1+an)/2S(2n)/Sn=2(a1+a2n)/(a1+an)=4a1+a2n=2a1+2ana2n=an+nd,其中d为