若log2(9x)+log2(x-1/3)=1,则lim(1+x+x^2+……+x^n)=?

2个回答

  • 由对数的运算法则可知:log2(9x) + log2(x-1/3) = log2[9x(x-1/3)] .

    又根据对数的性质可知,1可以写成:1 = log2(2)

    从而log2[9x(x-1/3)] = log2(2) 即:9x(x-1/3) = 2

    9x^2 - 3x -2 = 0

    十字交叉:(3x-2)(3x+1) = 0

    x1=2/3 x2= -1 (有两个解就有2种情况)

    由等比级数求和的公式可知:1+x+x^2+...+x^n = x^0 + x^1 +x^2 + ...+x^n

    = {1 * [ 1 - x^(n+1) ]} / (1-x)

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

    ( 情况1):当x1=2/3时,计算得 [1-x^(n+1)]/(1-x) = (1/3)^n,

    于是,当n趋于无穷大,lim(1/3)^n = 0

    (情况2):当x2=-1时,计算得 [1-x^(n+1)]/(1-x) = [1- (-1)^(n+1)]/2,

    于是,对n的奇偶性讨论:

    当n取偶数时,lim { [1- (-1)^(n+1)]/2 } = [1- (-1)]/2 = 1

    当n取奇数时,lim { [1- (-1)^(n+1)]/2 } = (1-1)/2 = 0