Question

第十二题是什么计算？

Answer 1

Sn = n²*An
Sn-1 = (n-1)²*An-1
因为：
An = Sn - Sn-1 = n²*An - (n-1)²*An-1
n²*An - An = (n-1)² * An-1
(n²-1)An = (n-1)² * An-1
(n+1)(n-1)*An = (n-1)² * An-1
(n+1)An = (n-1)An-1
An/An-1 = (n-1)/(n+1)
同理：
An-1/An-2 = (n-2)/n
An-2/An-3 = (n-3)/(n-1)
An-3/An-4 = (n-4)/(n-2)
……
A4/A3 = 3/5
A3/A2 = 2/4
A2/A1 = 1/3
这些等式两边分别相乘，并化简可以得到：
An/A1 = [1×2×3×……×(n-1)]/[3×4×……×(n+1)]
= (1×2)/[n(n+1)]
所以：
An = A1×2/[n(n+1)] = 2/[n(n+1)] = 2/n - 2/(n+1)