关于递推关系3 5 6 7

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  • (2)

    a[2] / a[1] = 1 / 3

    a[3] / a[2] = 2 / 4

    a[4] / a[3] = 3 / 5

    ...

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

    a[n] / a[n - 1] = (n - 1) / (n + 1)

    将左边全部乘起来再乘上a[1]得

    a[1] * (a[2] / a[1]) * ...* (a[n] / a[n - 1])

    =a[1] / a[1] * a[2] / a[2] * ...* a[n - 1] / a[n - 1] * a[n]

    =a[n] = 2 * (1 / 3) * (2 / 4) * (3 / 5) ...* (n - 2) / n * (n - 1) / (n + 1)

    = 1 * 2 * 2 * 3 / 3 * 4 / 4 * 5 / 5 ...* (n - 1) / (n - 1) / n / (n + 1)

    = 2 * 2 / n / (n + 1)

    = 4 / [(n + 1) * n]

    (3) //sqrt 表示根号下

    1 / [sqrt(n + 1) + sqrt(n)]

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

    =[sqrt(n + 1) - sqrt(n)] / [(n + 1) - n]

    =[sqrt(n + 1) - sqrt(n)] / 1

    =sqrt(n + 1) - sqrt(n)

    所以 a[n] - a[n - 1] = sqrt(n + 1) - sqrt(n)

    a[2] - a[1] = sqrt3 - sqrt2

    a[3] - a[2] = sqrt4 - sqrt3

    ...

    a[n] - a[n - 1] = sqrt(n + 1) - sqrt(n)

    左边全部加起来

    a[n] - a[n - 1] + a[n - 1] + a[n - 2] + ...+a[3] - a[2] + a[2] - a[1]

    =a[n] - a[1] = sqrt(n + 1) - sqrt(2)

    a[n] = sqrt(n + 1) - sqrt(2) + a[1] = sqrt(n + 1) - sqrt(2) + 1

    (5)

    因为 b[n + m] = b[n] * b[m]

    b[2] = b[1] * b[1] = 2^2

    b[3] = b[1] * b[2] = 2^3

    所以 b[n] = b[n - 1] * b[1] = b[n - 2] * b[1]^2 = ...= b[1] * b[1]^(n - 1)

    = b[1]^n = 2^n

    (6)

    a[n] - 2a[n - 1] = 1

    2(a[n - 1] - 2a[n - 2]) = 2

    2^2 * (a[n - 2] - 2a[n - 3]) = 2^2

    2^3 * (a[n - 3] - 2a[n - 4]) = 2^3

    .

    2^(n - 2) * (a[2] - 2a[1]) = 2^(n - 2)

    左边加起来得

    a[n] - 2^(n - 1) * a[1] = 2^0 + 2^1 + 2^2 ...+ 2^(n - 2) = 2^(n - 1) - 1

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

    (7)

    由 a[n + 1] - a[n] = 2^n

    所以

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

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

    ...

    a[2] - a[1] = 2^1

    左边加起来

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

    a[n] = 2^n - 1 - 1 + a[1] = 2^n

    S[n] = 2^1 + 2^2 + 2^3 + ...+ 2^n

    = 1 - 1 + 2^1 + 2^2 + 2^3 ...+ 2^n

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

    = 2^(n + 1) - 2