已知Sn=n²,求证1 S1 1 S2 -- 1 Sn>n n 1
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![已知Sn=n²,求证1 S1 1 S2 -- 1 Sn>n n 1](/uploads/image/f/4225192-16-2.jpg?t=%E5%B7%B2%E7%9F%A5Sn%3Dn%C2%B2%2C%E6%B1%82%E8%AF%811+S1+1+S2+--+1+Sn%3En+n+1)
由题,只要证明1/2+.+1/2^n>n/2(n>=2)用数学归纳法当n=2时,左边=1/2+1/3+1/4=13/12.右边=2/2=1,左边>右边,成立假设当n=m是时成立,即1/2+.+1/2^
Sn=2(1-3^n)/(1-3)=3^n-1S(n+1)=3*3^n-1S(n+1)/Sn=(3*3^n-1)/(3^n-1)=(3*3^n-3+2)/(3^n-1)=3+2/(3^n-1)(3n+
a1=1+根号2S3=9+3根号2a2=3+根号2an=2n-1+根号2sn=n^+n根号2bn=sn/n=n+根号假设bn,bk,bm成等比则m+n=2k,mn=k^2解得m=n,不符题意
an+1=Sn+3^nS(n+1)=S(n)+a(n+1)=2Sn+3^nS(n+1)-3^(n+1)=2[s(n)-3^n]即b(n+1)=2b(n)bn为等比数列,公比为2b1=S1-3^1=a1
Sn+1—Sn=an+1=Sn—n+3,即Sn+1=2Sn-n+3,所以Sn+1-(n+1)+2=2(Sn-n+2)又S1-1+2=3,所以Sn-n+2=3*2^n-1,所以bn=n/(3*2^n-1
1.na(n+1)=n[S(n+1)-Sn]=(n+2)SnnS(n+1)=2(n+1)SnS(n+1)/(n+1)=2*Sn/n所以{Sn/n}是公比为2的等比数列2.S1/1=a1=1所以Sn/n
an+2Sn*Sn-1=0其中an=Sn-Sn-1代入上式:Sn-Sn-1+2Sn*Sn-1=0a1=1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得:1/Sn-1-1/Sn+2=0即:1
由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn
第一个搞定我就不罗嗦了即1/Sn-1/Sn-1=2所以有1/Sn-1/Sn-1=21/Sn-1-1/Sn-2=21/Sn-2-1/Sn-3=2…………1/S2-1/S1=2叠加得1/Sn-1/S1=2
证明:(1)注意到:a(n+1)=S(n+1)-S(n)代入已知第二条式子得:S(n+1)-S(n)=S(n)*(n+2)/nnS(n+1)-nS(n)=S(n)*(n+2)nS(n+1)=S(n)*
1)证明:na[n+1]=(n+2)S[n]n(S[n+1]-S[n])=(n+2)S[n]nS[n+1]=2(n+1)S[n]S[n+1]/(n+1)=2*S[n]/n,(首项=S[1]/1=a[1
因为an=Sn-S(n-1),注意到(n-1)/[n(n+1)]=(n-1)/n-(n-1)/(n+1)=1-1/n-(1-2/(n+1))=2/(n+1)-1/n所以Sn+an=Sn+(Sn-S(n
an+1=Sn+1-Sn=[(n+2)/n]SnSn+1=[(2n+2)/n]Sn[Sn+1/(n+1)]/(Sn/n)=2所以,{Sn/n}是公比为2的等比数列.
证明:A(n+1)=Sn+3n+1,则An=S(n-1)+3n-2两式想减得A(n+1)-An=Sn+3n+1-(S(n-1)+3n-2)=An+3即A(n+1)+3=2(An+3)即(A(n+1)+
1/a^n=2^n/(2^2n-1)=1/(2^n+1)+1/(2^2n-1),因为1/(2^n+1)再问:恕我愚笨--可是能否告知1/(2^2n-2)如何用等比数列求和方式求,谢谢!再答:其实你可以
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
2an-2^n=sn2a(n-1)-2^(n-1)=s(n-1)两式想减,有2an-2a(n-1)-2^n+2^(n-1)=an2an-2a(n-1)-2^(n-1)-an=0an-2a(n-1)=2
n>1时sn=an(1-2/sn)=(sn-s(n-1))(1-2/sn)=sn-s(n-1)-2+2s(n-1)/sn整理可得:sn*s(n-1)=2(s(n-1)-sn)1/sn-1/s(n-1)
由Sn=n^2+3n得S(n-1)=(n-1)^2+3(n-1),两式相减,考虑到Sn-S(n-1)=an得an=2n-1+3=2n+4,于是得a(n-1)=2(n-1)+4,两式相减得an-a(n-