设{an}为等差数列,bn=(1/2)^an ,b3b4b5=512,b1b5+2b3b5+b3b7=400,则{bn}

4个回答

  • 3b4b5=512

    (1/2)^a3*(1/2)^a4*(1/2)^a5*=512

    (1/2)^(a3+a4+a5)=512

    (1/2)^(3a4)=512

    (1/2)^(3a4)=2^9

    (1/2)^(3a4)=(1/2)^(-9)

    3a4=-9

    a4=-3

    b4=(1/2)^a4

    =(1/2)^-3

    =2^3

    =8

    b1b5+2b3b5+b3b7=400

    (b3)^2+2b3b5+(b5)^2=400

    (b3+b5)^2=400

    (b3+b5)^2=400

    (b4/q+b4q)^2=400

    (b4/q+b4q)^2=400

    (8/q+8q)^2=400

    64(1/q+q)^2=400

    4(1/q+q)^2=25

    4/q^2+8+4q^2=25

    4/q^2-17+4q^2=0

    4q^4-17q^2+4=0

    (4q^2-1)(q^2-4)=0

    (2q-1)(2q+1)(q-2)(q+2)=0

    q=1/2或q=-1/2或q=2或q=-2

    b3b4b5=512

    b1q^2*b1q^3*b1^4=512

    (b1)^3*q^9=512

    当q=1/2时

    (b1)^3*(1/2)^9=512

    (b1)^3*(1/2)^9=(1/2)^-9

    (b1)^3*=(1/2)^-18

    b1=(1/2)^-6

    sn=b1(1-q^n)/(1-q)

    =(1/2)^(-6)*[1-(1/2)^n]/(1-1/2)

    =(1/2)^(-6)*[1-(1/2)^n]/(1/2)

    =(1/2)^(-7)*[1-(1/2)^n]

    =(1/2)^(-7)-(1/2)^(n-7)

    =128-(1/2)^(n-7)

    当q=-1/2时

    (b1)^3*(-1/2)^9=512

    (b1)^3*(-1/2)^9=(1/2)^(-9)

    (b1)^3*=-(1/2)^(-18)

    b1=-(1/2)^(-6)

    sn=b1(1-q^n)/(1-q)

    =-(1/2)^(-6)*[1-(-1/2)^n]/(1+1/2)

    =-(1/2)^(-6)*[1-(-1/2)^n]/(3/2)

    =-(1/2)^(-7)*[1-(-1/2)^n]/3

    =-(1/2)^(-7)+(-1/2)^(n-7)

    =-128+(-1/2)^(n-7)

    当q=2时

    (b1)^3*2^9=512

    (b1)^3*2^9=2^9

    (b1)^3=1

    b1=1

    sn=b1(1-q^n)/(1-q)

    =[1-2^n]/(1-2)

    =2^n-1

    当q=-2时

    (b1)^3*(-2)^9=512

    -(b1)^3*2^9=2^9

    (b1)^3=-1

    b1=-1

    sn=b1(1-q^n)/(1-q)

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

    =[(-2)^n-1]/3