>>488
n人が待っている確率を　P(n) とすると P(n)=ρ^n*P0 　( P0=P(0) )
1=納k=0,N]p(k)=P0{1-ρ^(N+1)}/(1-ρ) より　P0=(1-ρ)/{1-ρ^(N+1)}
L=納k=0,N]kp(k)=P0納k=0,N]kρ^k
ρL=P0納k=1,N+1](k-1)ρ^k との差をとって
(1-ρ)L=P0納k=1,N]ρ^k - P0Nρ^(N+1)
L={P0/(1-ρ)}納k=1,N]ρ^k - P0Nρ^(N+1)/(1-ρ)
={P0/(1-ρ)}*ρ(1-ρ^N)/(1-ρ) - P0Nρ^(N+1)/(1-ρ)
={ρ(1-ρ^N)-(1-ρ)Nρ^(N+1)} / [(1-ρ){1-ρ^(N+1)}]
=ρ{(1-ρ^N)-(1-ρ)Nρ^N} / [(1-ρ){1-ρ^(N+1)}]
=ρ{1-(N+1)ρ^N + Nρ^(N+1)} / [(1-ρ){1-ρ^(N+1)}]

Lq=納k=1,N](k-1)P(k)
=P0納k=1,N](k-1)ρ^k
=P0納k=1,N]kρ^k - P0納k=1,N]ρ^k
=L - P0*ρ(1-ρ^N)/(1-ρ)
=L - {ρ-ρ^(N+1)}/{1-ρ^(N+1)}
=L - [1-(1-ρ)/{1-ρ^(N+1)}]
=L - (1-P0)