已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=(3n^2)an+S(n-1)^2已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=3n^2*an+S(n-1)^2,an不等于0,n=2,3,4…(1)证明:数列{
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![已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=(3n^2)an+S(n-1)^2已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=3n^2*an+S(n-1)^2,an不等于0,n=2,3,4…(1)证明:数列{](/uploads/image/z/634619-11-9.jpg?t=%E5%B7%B2%E7%9F%A5An%28an%2Cbn%29%E6%98%AF%E6%9B%B2%E7%BA%BFy%3De%5Ex%E4%B8%8A%E7%9A%84%E7%82%B9%2Ca1%3Da%2CSn%E6%98%AF%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E4%B8%94%E6%BB%A1%E8%B6%B3Sn%5E2%3D%283n%5E2%29an%2BS%28n-1%29%5E2%E5%B7%B2%E7%9F%A5An%28an%2Cbn%29%E6%98%AF%E6%9B%B2%E7%BA%BFy%3De%5Ex%E4%B8%8A%E7%9A%84%E7%82%B9%2Ca1%3Da%2CSn%E6%98%AF%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E4%B8%94%E6%BB%A1%E8%B6%B3Sn%5E2%3D3n%5E2%2Aan%2BS%28n-1%29%5E2%2Can%E4%B8%8D%E7%AD%89%E4%BA%8E0%2Cn%3D2%2C3%2C4%E2%80%A6%281%29%E8%AF%81%E6%98%8E%EF%BC%9A%E6%95%B0%E5%88%97%7B)
已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=(3n^2)an+S(n-1)^2已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=3n^2*an+S(n-1)^2,an不等于0,n=2,3,4…(1)证明:数列{
已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=(3n^2)an+S(n-1)^2
已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=3n^2*an+S(n-1)^2,an不等于0,n=2,3,4…
(1)证明:数列{b(n+2)/bn}(n>=2)是常数数列
(2)确定a的取值集合M,使a属于M时,数列{an}是单调递增数列
已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=(3n^2)an+S(n-1)^2已知An(an,bn)是曲线y=e^x上的点,a1=a,Sn是数列{an}的前n项和,且满足Sn^2=3n^2*an+S(n-1)^2,an不等于0,n=2,3,4…(1)证明:数列{
(1)证明:b(n+2)/bn=e^a(n+2)/e^an=e^[a(n+2)-an]
要证明{b(n+2)/bn}为常数数列,只需证a(n+2)-an为常数;
∵Sn^2=3n^2*an+S(n-1)^2
∴Sn^2-S(n-1)^2=[Sn+S(n-1)][Sn-S(n-1)]=[Sn+S(n-1)]*an=3n^2*an
∴Sn+S(n-1)=3n^2……①,S(n+1)+Sn=3(n+1)^2……②,S(n+2)+S(n+1)=3(n+2)^2……③
②-①:S(n+1)-S(n-1)=6n+3,即a(n+1)+an=6n+3……④
③-②:S(n+2)-Sn=6n+9,即a(n+2)+a(n+1)=6n+9……⑤
⑤-④:a(n+2)-an=6n+9-6n-3=6……⑥
∴a(n+2)-an=6为常数,数列{b(n+2)/bn}为常数列,且b(n+2)/bn=e^6
(2)由①:S2+S1=12,a2+2a1=12
∴a2=12-2a1=12-2a;由④:a3+a2=15,a4+a3=21
∴a3=15-a2=3+2a;a4=21-a3=18-2a
由⑥可看出:数列{a2k}、{a(2k+1)}(k属于Z+)分别是以a2、a3为首项,6为公差的等差数列
∴a2k=a2+6(k-1),a(2k+1)=a3+6(k-1),a(2k+2)=a4+6(k-1)(k属于N*)
数列{an}为单调递增数列a1<a2且a2k<a(2k+1)<a(2k+2)对任意k属于N*成立
a1<a2且a2+6(k-1)<a3+6(k-1)<a4+6(k-1)a1<a2<a3<a4
a1<12-2a<3+2a<18-2a9/4<a<15/4
∴M={a|9/4<a<15/4}