求解下列组合数求和问题.实际上只需要求得一式的表达式二式的也可求出,但我无法求出一式的和,求大神帮忙解决

先求二式再求一式比较方便.
用C(m,k)表示m中选k的组合数,并约定k < 0或k > m时C(m,k) = 0.
首先,有组合恒等式∑{0 ≤ i ≤ k} C(m,i)C(n,k-i) = C(m+n,k) ①.
证明只需考虑恒等式(1+x)^m·(1+x)^n = (1+x)^(m+n)两边的k次项系数.
∵k·C(2n,k) = 2n·C(2n-1,k-1),
∴∑{0 ≤ k ≤ n-1} k²·C(2n,k)²
= 4n²·∑{1 ≤ k ≤ n-1} C(2n-1,k-1)²
= 4n²·∑{0 ≤ k ≤ n-2} C(2n-1,k)².
而由①,C(4n-2,2n-1) = ∑{0 ≤ k ≤ 2n-1} C(2n-1,k)C(2n-1,2n-1-k)
= ∑{0 ≤ k ≤ 2n-1} C(2n-1,k)²
= 2·C(2n-1,n-1)²+ ∑{0 ≤ k ≤ n-2} C(2n-1,k)² + ∑{n+1 ≤ k ≤ 2n-1} C(2n-1,k)²
= 2·C(2n-1,n-1)²+ ∑{0 ≤ k ≤ n-2} C(2n-1,k)² + ∑{n+1 ≤ k ≤ 2n-1} C(2n-1,2n-1-k)²
= 2·C(2n-1,n-1)²+ ∑{0 ≤ k ≤ n-2} C(2n-1,k)² + ∑{0 ≤ k ≤ n-2} C(2n-1,k)²
= 2·C(2n-1,n-1)²+ 2·∑{0 ≤ k ≤ n-2} C(2n-1,k)²,
∴∑{0 ≤ k ≤ n-2} C(2n-1,k-1)² = C(4n-2,2n-1)/2-C(2n-1,n-1)²,
∴∑{0 ≤ k ≤ n-1} k²·C(2n,k)² = 2n²·C(4n-2,2n-1)-4n²·C(2n-1,n-1)² ②.
∵k·C(2n,k) = 2n·C(2n-1,k-1),(2n-k)·C(2n,k) = 2n·C(2n-1,k),
∴∑{0 ≤ k ≤ n-1} k(2n-k)·C(2n,k)²
= 4n²·∑{1 ≤ k ≤ n-1} C(2n-1,k-1)C(2n-1,k)
= 4n²·∑{0 ≤ k ≤ n-2} C(2n-1,k)C(2n-1,k+1).
而由①,C(4n-2,2n-2) = ∑{0 ≤ k ≤ 2n-2} C(2n-1,k)C(2n-1,2n-2-k)
= ∑{0 ≤ k ≤ 2n-2} C(2n-1,k)C(2n-1,k+1)
= C(2n-1,n-1)²+ ∑{0 ≤ k ≤ n-2} C(2n-1,k)C(2n-1,k+1) + ∑{n ≤ k ≤ 2n-2} C(2n-1,k)C(2n-1,k+1)
= C(2n-1,n-1)²+ ∑{0 ≤ k ≤ n-2} C(2n-1,k)C(2n-1,k+1) + ∑{0 ≤ k ≤ n-2} C(2n-1,k+1)C(2n-1,k)
= C(2n-1,n-1)²+ 2·∑{0 ≤ k ≤ n-2} C(2n-1,k)C(2n-1,k+1),
∴∑{0 ≤ k ≤ n-2} C(2n-1,k)C(2n-1,k+1) = C(4n-2,2n-2)/2-C(2n-1,n-1)²/2,
∴∑{0 ≤ k ≤ n-1} k(2n-k)·C(2n,k)² = 2n²·C(4n-2,2n-2)-2n²·C(2n-1,n-1)² ③.
②+③得2n·∑{0 ≤ k ≤ n-1} k·C(2n,k)²
= 2n²·(C(4n-2,2n-1)+C(4n-2,2n-2))-6n²·C(2n-1,n-1)²
= 2n²·C(4n-1,2n-1)-6n²·C(2n-1,n-1)²,
∴∑{0 ≤ k ≤ n-1} k·C(2n,k)² = n·C(4n-1,2n-1)-3n·C(2n-1,n-1)².

这个好象没什么公式可以看的