若数列是等比数列,前n项和Sn=2^n-1
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/14 04:11:36
A(n+1)=2S(n)+1,A(n)=2S(n-1)+1,A(n+1)-A(n)=2[S(n)-S(n-1)]=2[A(n)],A(n+1)=3A(n)所以,数列{A(n)}是首项为1,公比为3的等
由题意可得a1=3-a,a2=S2-S1=6,a3=S3-S2=18,由等比数列可得36=(3-a)•18,解得a=1,故选:B
∵Sn=n-an,∴a(n+1)=S(n+1)-S(n)=(n+1)-a(n+1)-n+a(n)=1+a(n)-a(n+1);∴2a(n+1)=1+a(n);∴2a(n+1)-2=1+a(n)-2,即
n=1/anan=q的n-1次方bn=q的1-n次方bn=1+1/q+1/q²+…1/q的n-1次方bn的前n项和=(1-(1/q)的n次)/(1-1/q)
设公比为q,当q=-1时,等比数列{an}的各项是a,-a,a,-a,a,-a…的形式,a≠0.又已知Sn是实数等比数列{an}前n项和,故当n为偶数时,Sn=0,当n为奇数时,Sn=a,故选D.
设等比数列{a[n]}的公比为q则S[n]=a[1](1-qⁿ)/(1-q)=2(1-qⁿ)/(1-q)则S[n]+1=2(1-qⁿ)/(1-q)+1S[1]+1=
an+sn=-2n-1,当n=1时,a1+s1=-3,则a1=-3/2.由已知得:sn=-2n-1-an当n大于或等于2时,则an=sn-s(n-1)=-2n-1-an-[-2(n-1)-1-a(n-
Sn=n-5an-85S1=1-5a1-85即a1=1-5a1-85解得a1=-14an=Sn-S(n-1)=n-5an-85-[(n-1)-5a(n-1)-85]=-5an+5a(n-1)+16an
证:(1)根号Sn+1=(a1+1)*2^(n-1)=4*2^(n-1)=2^(n+1)Sn+1=2^(2n+2)=4^(n+1).1Sn=4^n.21式-2式Sn+1-Sn=4^(n+1)-4^na
(1)∵原命题是数列{an}的首项a1=5,前n项和为Sn,若Sn+1=2Sn+n+5(n∈N*),则数列{an+1}是等比数列;∴它的逆命题是数列{an}的首项a1=5,前n项和为Sn,若数列{an
∵a(n+1)=(n+2)Sn/n且a(n+1)=S(n+1)-Sn∴S(n+1)-Sn=(n+2)*Sn/n∴S(n+1)=[(n+2)/n+1]Sn=(2n+2)/n*Sn∴S(n+1)/(n+1
Sn=2an-3n+5S(n-1)=2a(n-1)-3(n-1)+5相减an=2a(n-1)+3an+3=2a(n-1)+6an+3=2[2a(n-1)+3]
为了避免混淆,我把下角标放在内.首先从数列本身的基本意义出发a=S-S其次,从已知a=S(n+2)/n出发a=S*(n+1)/(n-1)因此S-S=S*(n+1)/(n-1)移项整理S=S
1、A(n+1)=(n+2)sn/n=S(n+1)-Sn即nS(n+1)-nSn=(n+2)SnnS(n+1)=(n+2)Sn+nSnnS(n+1)=(2n+2)SnS(n+1)/(n+1)=2Sn/
Sn=a^n-1的a是a的n次方?1.a=1时,Sn=0,an=0,是常数列;2.a不等于1时,a1=S1=a-1,S(n-1)=a^(n-1)-1,an=Sn-S(n-1)=a^n-a^(n-1)=
Sn=4An-3S(n-1)=4A(n-1)-3Sn-S(n-1)=An=4An-3-[4A(n-1)-3]=4an-3-4A(n-1)+3=4An-4A(n-1)3An=4A(n-1)An/A(n-
1.证:Sn=(3an-n)/2Sn-1=[3a(n-1)-(n-1)]/2an=Sn-Sn-1=[3an-3a(n-1)-1]/2an=3a(n-1)+1an+1/2=3a(n-1)+3/2=3[a
Sn+an=n^2+3n+5/2①当n=1时,S1+a1=1^2+3*1+5/2=13/2而S1=a1,所以2a1=13/2,即a1=13/4,所以a1-1=9/4;又S(n-1)+a(n-1)=(n
设公比为q,因为a1=1,即:a(n)=q^(n-1)则:S(n)=(1-q^n)/(1-q)若{Sn}为等差数列,设公差为d则:S(n)=S(n-1)+d即:d=S(n)-S(n-1)=(1-q^n
a1=S1=m+3a2=S2-a1=m+9-(m+3)=6a3=S3-a1-a2=27+m-6-(m+3)=18数列是等比数列q=a3/a2=18/6=3a1=a2/q=6/3=2m+3=2m=-1也