Question

已知p²-p-1=0,1-q-q²=0，且pq≠1，求（pq+1）/q的值

Answer 1

1－q－q²=0即为q²－q－1=0，也就是说p、q是方程x²－x－1=0的根，从而两根之积为pq=－1，所以(pq＋1)/q=0。

Answer 2

P(ξ=1)+P(ξ=2)+P(ξ=3)=a+a/2+a/3=1 a=6/11
P(n= -1)+P(n= -2)+P(n= -3)=b+b/4+b/9=1 b=36/49

P(ξ=k,n=-m)=6/11k*36/(49s^2)=216/(539ks^2) (k=1,2,3 s=1,2,3)
P(1,-1)=ab=216/539 P(1,-2)=ab/4=54/539 P(1,-3)=ab/9=24/539
P(2,-1)=ab/2=108/539 P(2,-2)=ab/8=27/539 P(2,-3)=ab/18=12/539
P(3,-1)=ab/3=72/539 P(3,-2)=ab/12=18/539 P(3,-3)=ab/27=8/539

P(ξ＋n=-2)=P(1,-3)=ab/9=24/539
P(ξ＋n=-1)=P(2,-3)+P(1,-2)=ab/4=54/539
P(ξ＋n=0)=P(1,-1)+P(2,-2)+P(3,-3)=251/539
P(ξ＋n=1)=P(2,-1)+P(3,-2)=126/539
P(ξ＋n=2)=P(3,-1)=ab/3=72/539