已知数列{an}的前n项和Sn=a(Sn-an+1)(a为常数,a>0),且4a3是a1与2a的等差中项?急

3个回答

  • 由题意得:Sn=a(Sn-an+1)(a>0)

    则易得:Sn=[a/(a-1)]*(an-1)

    根据:an=Sn-S(n-1)(n>1)

    所以:an=[a/(a-1)]*(an-1)-[a/(a-1)]*[a(n-1)-1]=[a/(a-1)]*[an-a(n-1)]

    得:an/a(n-1)=a

    因此:数列{an}是以a为公比的等比数列

    an=a1*a^(n-1)

    又因为:S1=a(S1-a1+1)且S1=a1

    即:a1=a

    所以:an=a^n

    4a3是a1与2a的等差中项,则有8a3=a1+2a

    即有8a^3=a+2a=3a

    a^2=3/8

    a=根号6/4

    故an=(根号6/4)^n

    bn=2n+1/an=2n+(4/根号6)^n

    Tn=2(1+n)n/2+4/根号6*((4/根号6)^n-1)/(4/根号6-1)

    =n(n+1)+4((4/根号6)^n-1)/(4-根号6)