- 92 名前：１３２人目の素数さん mailto:sage [2007/12/29(土) 15:17:17 ]
- >>66
C[2n,k]･k = 2n･C[2n-1,k-1] (1≦k≦2n), >>73
(与式)/n = (1/n)Σ[k=1,n-1] C[2n,k]･k = 2Σ[k=1,n-1] C[2n-1,k-1]
= Σ[k'=0,n-2] C[2n-1,k'] + Σ[L=n+1,2n-1] C[2n-1,L] (k'=k-1, L=2n-k)
= (1+1)^(2n-1) - C[2n-1,n-1] - C[2n-1,n]
= 2^(2n-1) - C[2n,n],
念のため。
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