求（2n*2n)/(2n+1)(2n-1)的前n项和

求（2n*2n)/(2n+1)(2n-1)的前n项和
求（2n*2n)/(2n+1)(2n-1)的前n项和

求（2n*2n)/(2n+1)(2n-1)的前n项和

4n^2 / ((2n+1)(2n-1))
= 2n^2 * (1/(2n-1) - 1/(2n+1))
= (2n^2 / (2n-1)) - (2n^2 / (2n+1))
我们分别求 2n^2 / (2n-1) 和 2n^2 / (2n+1) 的前n项和。
Σ 2n^2 / (2n-1)
= Σ (n(2n-1) + (1/2)(2n-1) + (1/...

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4n^2 / ((2n+1)(2n-1))
= 2n^2 * (1/(2n-1) - 1/(2n+1))
= (2n^2 / (2n-1)) - (2n^2 / (2n+1))
我们分别求 2n^2 / (2n-1) 和 2n^2 / (2n+1) 的前n项和。
Σ 2n^2 / (2n-1)
= Σ (n(2n-1) + (1/2)(2n-1) + (1/2)) / (2n-1)
= Σ (n + (1/2) + (1/2)/(2n-1))
= n(n+1)/2 + n/2 + Σ (1/2)/(2n-1)
Σ 2n^2 / (2n+1)
= Σ (n(2n+1) - (1/2)(2n+1) + (1/2)) / (2n+1)
= Σ (n - (1/2) + (1/2)/(2n+1))
= n(n+1)/2 - n/2 + Σ (1/2)/(2n+1)
所以，二式相减：
原式 = n + Σ (1/2)/(2n-1) - Σ (1/2)/(2n+1)
= n + (1/2)(1+1/3+1/5+...+1/(2n-1)) - (1/2)(1/3+1/5+...+1/(2n-1)+1/(2n+1))
= n + (1/2)(1 - 1/(2n+1))
= n + n / (2n+1)
= 2n(n+1) / (2n+1)

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