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)},
即
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.