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_数学解答

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

人气：297 ℃　时间：2019-08-21 06:33:39

解答

(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…...