04/10/22 19:45:59
>>789
できた。
Σ[k=1,n] (k^2)C[2n,n-k]
=Σ[k=0,n] (k^2)C[2n,n-k]
=Σ[k=0,n] ((n-k)^2)C[2n,k]
=(1/2)Σ[k=0,2n] ((n-k)^2)C[2n,k]
=(1/2)Σ[k=0,2n] (n-k)(n-k-1)C[2n,k]
=(1/2)(Σ[k=0,2n] C[2n,k]t^(n-k))''|t=1
=(1/2)(t^n(1+1/t)^(2n))''|t=1
=(1/2)((t+2+1/t)^n)''|t=1
=(1/2)(n(n-1)(t+2+1/t)^(n-2)(1-1/t^2)+n(t+2+1/t)^(n-1)(2/t^2)))|t=1
=n*(4^(n-1))