设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an ps:只需第三问!须详述!

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  • a(n)=aq^(n-1),

    b(n) = na(1)+(n-1)a(2)+...+2a(n-1)+a(n),

    m=b(1)=a(1)=a,

    3m/2 = b(2) = 2a(1)+a(2) = 2m + a(2), a(2)=-m/2.

    q = a(2)/a(1)=-1/2

    a(n) = m*(-1/2)^(n-1).

    m=1时,a(n) = (-1/2)^(n-1).

    b(n) = n*1 + (n-1)(-1/2) + (n-2)(-1/2)^2 + ... + 2(-1/2)^(n-2) + (-1/2)^(n-1),

    -2b(n) = n*(-2) + (n-1)*1 + (n-2)(-1/2) + ... + 2(-1/2)^(n-3) + (-1/2)^(n-2),

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

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

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

    b(n) = 2n/3 + 2[1 - (-1/2)^n]/9

    a(n) = m(-1/2)^(n-1),m不为0.

    s(n) = a(1)+a(2)+...+a(n) = m[1+(-1/2)+...+(-1/2)^(n-1)] = m[1 - (-1/2)^n]/[1-(-1/2)]

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

    s(2n-1) = (2m/3) - (2m/3)(-1/2)^(2n-1) = 2m/3 + (2m/3)(1/2)^(2n-1), 单调递减. 2m/3 s(2n) >= m/2.

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