已知p²-p-1=0,1-q-q²=0,且pq≠1,求(pq+1)/q的值

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已知p²-p-1=0,1-q-q²=0,且pq≠1,求(pq+1)/q的值已知p²-p-1=0,1-q-q²=0,且pq≠1,求(pq+1)/q的值已知p&su

已知p²-p-1=0,1-q-q²=0,且pq≠1,求(pq+1)/q的值
已知p²-p-1=0,1-q-q²=0,且pq≠1,求(pq+1)/q的值

已知p²-p-1=0,1-q-q²=0,且pq≠1,求(pq+1)/q的值
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.

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=...

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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

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