设蠕虫走n次后距终点的距离为a(n), 则a(n)=[(n+1)/n][a(n-1)-0.01], a(0)=1, 即:
a(0)=1
a(1)=(2/1)[a(0)-0.01]
a(2)=(3/2)[a(1)-0.01]
a(3)=(4/3)[a(2)-0.01]
...
a(n)=[(n+1)/n][a(n-1)-0.01]
所以:
a(0)=1
(1/2)a(1)=(1/2)(2/1)[a(0)-0.01]
(1/3)a(2)=(1/3)(3/2)[a(1)-0.01]
(1/4)a(3)=(1/4)(4/3)[a(2)-0.01]
...
[1/(n+1)]a(n)=[1/(n+1)][(n+1)/n][a(n-1)-0.01]
相加,得:
[1/(n+1)]a(n)=1-0.01[1/2+1/3+1/4+...+1/(n+1)]
所以,当1-0.01[1/2+1/3+1/4+...+1/(n+1)]≤0,即1/2+1/3+1/4+...+1/(n+1)≥100时,a(n)≤0,此时蠕虫达到终点.
估计n的值:
1/2+1/3+1/4+...+1/(n+1)≈100
1+1/2+1/3+1/4+...+1/(n+1)≈101
欧拉常数+ln(n+1)≈101
ln(n+1)≈100.42278
n≈4.10*10^43 |