X不停留在同一顶点处,每隔一秒向其他三个顶点以相同的概率移动,所以,
每个时刻,X在点Ai的概率等于X在上一个时刻从其余3点移动到Ai的概率.
(1)P2(1)=(1/3)P1(0)+(1/3)P3(0)+(1/3)P4(0)=(1/3)[P1(0)+P3(0)+P4(0)]=(1/3)[1-P2(0)]=1/6
(2)同样的,P2(N)=(1/3)P1(N-1)+(1/3)P3(N-1)+(1/3)P4(N-1)
=(1/3)[P1(N-1)+P3(N-1)+P4(N-1)]=(1/3)[1-P2(N-1)]
(3)P2(N)=(1/3)[1-P2(N-1)]=1/3-(1/3)P2(N-1)=1/3-(1/3)^2[1-P2(N-2)]
=1/3-(1/3)^2+(1/3)^2P2(N-2)=1/3-(1/3)^2+(1/3)^3[1-P2(N-3)]
=1/3-(1/3)^2+(1/3)^3-(1/3)^3P2(N-3)=.
=1/3-(1/3)^2+(1/3)^3-(1/3)^4+...+(-1)^N(1/3)^NP2(0)
=(1/3)[1-(-1/3)^N]/[1-(-1/3)]+(-1)^N(1/3)^N*(1/2)
=(1/4)[1+(-1/3)^N]
={(1/4)[1-1/3^N]N为奇数
{(1/4)[1+1/3^N]N为偶数
(4)同理,P1(N)=(1/3)[1-P1(N-1)]
所以,P1(N)
=1/3-(1/3)^2+(1/3)^3-(1/3)^4+...+(-1)^N(1/3)^NP1(0)
=(1/3)[1-(-1/3)^N]/[1-(-1/3)]+(-1)^N(1/3)^N*(1/4)
=1/3
高二整这个,有点难哦,老师尽变着法想着刁难学生哦.无良啊