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+
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/06/16 04:41:55
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,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
(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}
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}
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+
已知An(an,bn)是曲线y=(e)^x上的点,Sn是数列{an}的前n项和,并且满足an0,a1=a,(Sn)^2=
已知数列{an}的前n项和sn满足sn=an^2+bn,求证{an}是等差数列
已知数列{an}的前n项的和Sn,满足6Sn=an2+3an+2且an>0.(1)求首项a1;(2)证明{an}是
已知数列an满足;a1=1,an+1-an=1,数列bn的前n项和为sn,且sn+bn=2
已知数列{an}a1=2前n项和为Sn 且满足Sn Sn-1=3an 求数列{an}的通项公式an
已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证:{1/Sn}是等差数列
已知数列an的前n项和sn,且满足2sn+an=2,bn=2
已知数列{an}满足an>0且对一切n属于正整数,都有a1^3+a2^3+...+an^3=sn^2,sn是{an}的前
已知数列{an}的前n项和为Sn,满足2Sn+3=3an(n是正整数),{bn}是等差数列,且b2=a1,b4=a1+4
已知数列{an}的前n项和为Sn,且满足Sn=Sn-1/2Sn-1 +1,a1=2,求证{1/Sn}是等差数列
已知数列an满足a1=2 其前n项和为Sn Sn =n+7~3an 数列bn满足 bn=an~1 证明数列bn是等差数列