若an sn=n,cn=an-1,求证,cn是等比数列,并求an
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当n>1时,an=sn-sn-1代入化简得:1/Sn-1/Sn-1=2所以:1/Sn=2n-1所以:Sn=1/(2n-1)当n>1时,an=sn-sn-1=-2/[(2n-1)(2n-3)]当n=1时
2Sn²=2anSn-an,知an=Sn-S(n-1)代换后化简可得(过程不难但打起来很闹心……)1/Sn-1/S(n-1)=2故而1/Sn的通项公式为1/Sn=2n-1(1/S1=1),即
2sn^2=2ansn-an2sn^2=2[sn-s(n-1)]sn-[sn-s(n-1)]2sn^2=2sn^2-2sns(n-1)-sn+s(n-1)2sns(n-1)+sn-s(n-1)=01/
已知:数列{an}中,a1=1,Sn为数列{an}的前n项和,且2an/(anSn-Sn²)=1,(n≥2);(1).证明数列{1/Sn}是等差数列;(2).求数列{an}的通项公式.(1)
an=2n-1a(n+1)=2n+1a(n+1)-an=2已知cn=2^n,[a(n+1)-an][b(n+1)-bn]=cn,则b(n+1)-bn=2^(n-1)b(n+1)-2^n=bn+2^(n
lim{[(3n^2+cn+1)/(an^2+bn)]-4n}=5lim{[(3n^2+cn+1)-4n(an^2+bn)]/(an^2+bn)}=5lim{[-4an^3+(3-4b)n^2+cn+
an=Sn-Sn-1在往下可以得到1/Sn是一个等差数列
太简单了.2(S_n)^2=2a_nS_n-a_n=>2S_n(S_n-a_n)=-a_n=>2S_n*S_{n-1}=-a_n2S_n*S_{n-1}=-(S_n-S_{n-1})2=-1/S_{n
an=sn-s(n-1),2sn^2=2(sn-sn-1)sn-sn+s(n-1)=2sn^2-2s(n-1)sn-sn+s(n-1)2sns(n-1)=s(n-1)-sn2=1/sn-1/s(n-1
1.证:n≥2时,2Sn²=2anSn-an=2[Sn-S(n-1)]Sn-[Sn-S(n-1)]整理,得S(n-1)-Sn=2SnS(n-1)等式两边同除以SnS(n-1)1/Sn-1/S
(本小题满分12分)(Ⅰ)∵an+1=2an+1∴an+1+1=2(an+1),∵a1=1,a1+1=2≠0…(2分)∴数列{an+1}是首项为2,公比为2的等比数列.∴an+1=2×2n−1,∴an
n=1时,(s1-1)^2=s1*s1即-2s1+1=0解得s1=1/2n=2时,(s2-1)^2=(s2-s1)*s2解得:s2=2/3n=3时,(s3-1)^2=(s3-s2)*s3解得:s3=3
2(S_n)^2=2a_nS_n-a_n=>2S_n(S_n-a_n)=-a_n=>2S_n*S_{n-1}=-a_n2S_n*S_{n-1}=-(S_n-S_{n-1})2=-1/S_{n-1}+1
由题意知:2an/[anSn-(Sn)²]=1(n>1)则:(Sn)²-anSn+2an=0(n>1)又因为:an=Sn-S(n-1)(n>1)所以:(Sn)²-[Sn-
(1)a(n+1)-an=(n+1+2013)-(n+2013)=1∴b(n+1)-bn=cn/[a(n+1)-an]=cn=2^n+n∴bn-b(n-1)=2^(n-1)+n-1...b2-b1=2
证明:(1)∵an2-2anSn+1=0,an=Sn-Sn-1(n≥2)∴(Sn-Sn-1)2-2(Sn-Sn-1)Sn+1=0⇒Sn2-Sn-12=1故{Sn2}成等差数列.(2)∵a12-2a12
...这题和an有什么关系吗?cn=(2n-1)/2n,Tn=c1+c2+...+cn.当n=1时,Tn=1/2>-1/2成立.假设当n=k时也成立,即c1+c2+...+ck>-1/2√(k)①则n
Cn=an*bn=n*(4^n-1);Sn=C1+C2+C3+.+Cn;Sn=1*(4-1)+2*(4^2-1)+3*(4^3-1)+.+n*(4^n-1);所以Sn=4+2*4^2+3*4^3+.n