已知Sn=n^2 n,证明 1 3
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![已知Sn=n^2 n,证明 1 3](/uploads/image/f/4225189-13-9.jpg?t=%E5%B7%B2%E7%9F%A5Sn%3Dn%5E2+n%2C%E8%AF%81%E6%98%8E+1+3)
有条件a1=2,d=2吧,an=2n,S1=a1=1*(1+1),其满足,假设Sj=j^2+j=j(j+1),而a(j+1)=2(j+1),则S(j+1)=Sj+a(j+1)=(j+1)(j+2),满
证明an=Sn-S(n-1)=100n-n^2-[100(n-1)-(n-1)^2]=100n-n^2-[100n-100-(n^2-2n+1)]=100n-n^2-(-n^2+102n-101)=1
简单不就是放缩法Sn/Sn+1=(2^n-1)/(2^n+1-1)(Sn/Sn+1)-0.5=-1/(2^n+1-1)<0∴Sn/Sn+1<0.5则S1/S2+S2/S3+.+Sn/Sn+1<0.5+
证:由a1=1,an+1=[(n+2)/n]Sn(n=1,2,3)知a2=3a1S2/2=4a1/2=2S1/1=1∴(S2/2)/(S1/1)=2又an+1=Sn+1-Sn(n=1,2,3,…)则S
第一种方法:①an+1=Sn+1-Sn②an=Sn-Sn_1(n≥2)①-②得an+1-an=Sn+1+Sn_1-2Sn=(n+1)(a1+an+1)/2+(n-1)(an+an_1)/2-n(a1+
(1).看到Sn的式子,可以把An变为Sn-Sn-1,所以将原式变为Sn=n^2(Sn-Sn-1)-n(n-1).分解移项,得(n^2-1)Sn+n^2Sn-1+n(n-1)两边同除n(n-1)得(n
再问:设bn=Sn/n,求满足bn>2012/2013的最小正整数n.再答:
sn=n(2n-1)sn-1=(n-1)(2n-3)an=4n-3其实是一个等差数列再问:不懂。。。再答:对于一个数列,前N项满足该通式,则前N-1项也满足该式子做差即可,是高中常用的数列解决方法
1.证明:有:a(n+1)=S(n+1)-Sn=(n+2)Sn/n,整理:S(n+1)/(n+1)=2*Sn/n故:数列{Sn/n}是等比数列,其首项为1,公比为2.2.{an}可看成以下两等差数列构
Sn=1/1*2+1/2*3,...,1/n*(n+1)=(1-1/2)+(1/2-1/3)+.+[1/n-1/(n+1)]=1-1/(n+1)=n/(n+1)用数学归纳法证:当k=1时:S1=1/1
∵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
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
Sn=2n-an,(1)S(n+1)=2*(n+1)-a(n+1)(2)(2)-(1)得:a(n+1)=2-a(n+1)+an.即:2*a(n+1)=2+an.变形:2*[a(n+1)-2]=an-2
证:S2=4a1+2=4×1+2=6n≥2时,S(n+1)=4an+2=4[Sn-S(n-1)]+2S(n+1)-2Sn+2=2Sn-4S(n-1)+4[S(n+1)-2Sn+2]/[Sn-2S(n-
n>=2时:∵an=2Sn^2/[(2Sn)-1]∴Sn-(Sn-1)=2Sn^2/[(2Sn)-1]两边同时乘以(2Sn)-1并化简得2Sn(Sn-1)+Sn-(Sn-1)=0两边同时除以Sn(Sn
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
s(n+1)-sn=(n+2)/n*sns(n+1)/n+1=2sn/n所以sn/n是等比数列,公比为2,首项为1所以s(n+1)/n+1=1*2^n即s(n+1)=(n+1)*2^n=4(n+1)*
因为n,an,Sn成等差数列所以2an=Sn+n又因为an=Sn-Sn-1所以Sn+n=2Sn-1+2n左右两边同时加2Sn+n+2=2Sn-1+2n+2右边再变化Sn+n+2=2Sn-1+2n+2-
当n=1时,a1=S1=1当n≥2时,an=Sn-S(n-1)=3n²-2n-3(n-1)²+2(n-1)=6n-5∵当n=1时,满足an=6n-5又∵an-a(n-1)=6n-5