2、1、3、4、7、11从第3个数开始每个数是前面2个数之合求第2012个数是多少

3个回答

  • 令第n项为a[n],

    则a[n] = a[n - 1] + a[n - 2],

    设a[n] - ka[n - 1] = r {a[n - 1] - ka[n - 2]},

    解出r = 1/2 (1 - Sqrt[5]),k = 1/2 (1 + Sqrt[5]),

    令b[n] = a[n] - ka[n - 1],

    则b[n] = rb[n - 1] = r^(n - 4) b[4],

    即a[n] - ka[n - 1] = r^(n - 4) b[4],

    所以a[n]/r^n - k/r a[n - 1]/r^(n - 1) = r^-4 b[4],

    令c[n] = a[n]/r^n,

    c[n] - k/r c[n - 1] = r^-4 b[4],

    设x使得c[n] + x = k/r {c[n - 1] + x},

    解得x = (r^-4 b[4])/(k/r - 1),

    c[n] + x = (k/r)^(n - 4) {c[4] + x},

    所以

    a[n]/r^n + (r^-4 (a[4] - ka[3]))/(k/r - 1) = (k/r)^(

    n - 4) {a[4]/r^4 + (r^-4 (a[4] - ka[3]))/(k/r - 1)},

    a[n] = {(k/r)^(

    n - 4) {a[4]/r^4 + (r^-4 (a[4] - ka[3]))/(

    k/r - 1)} - (r^-4 (a[4] - ka[3]))/(k/r - 1)}*r^n,

    代入即得a[2012]=1.879942398662892*10^420.

    答案是用软件算出来的,有点吓人,也不知道对不对,不过你可以通过这种方法求出数列的通项公式.

    验算从第4项到第10项都是对的.