已知数列{an}是首项为a1=1/4,公比q=1/4的等比数列,设bn+2=3(log1/4)an(n∈N*),数列{C

2个回答

  • (1)由题意,可得

    an=(1/4)^n;

    那么:

    bn+2=3*log(1/4)an=3n;

    所以:

    bn=3n-2,为等差数列;

    (2)由条件Cn= an*bn得到:

    Cn= (1/4)^n*(3n-2)=3n*(1/4)^n-2*(1/4)^n

    记Cn的前n项和为Sn;那么:

    Sn=3[1/4+2*(1/4)^2+……+n*(1/4)^n]-2*(1/4+(1/4)^2+……+(1/4)^n);

    记Pn=1/4+2*(1/4)^2+……+n*(1/4)^n; --------(1)

    则有:

    1/4*Pn=(1/4)^2+2*(1/4)^3+……+n*(1/4)^(n+1); ------(2)

    (1)-(2)得到:

    3/4 Pn=1/4+(1/4)^2+(1/4)^3+……+(1/4)^n-n*(1/4)^(n+1)

    = 1/3*(1-(1/4)^n)- n*(1/4)^(n+1)

    所以Sn可变形为:

    Sn=3[1/3*(1-(1/4)^n)- n*(1/4)^(n+1)]-2*[1/3*(1-(1/4)^n)]

    =1/3*[1-(1/4)^n]-3n*(1/4)^(n+1);