Question

3.若a1=0,an+1=an+2(n-1),则an=？

Answer 1

因为：
An+1 = An + 2(n - 1)
那么也有：
An = An-1 + 2(n-1 - 1) = An-1 + 2n- 4
则有：
An - An-1 = 2n - 4
An-1 - An-2 = 2(n-1) - 4
An-2 - An-3 = 2(n-2) - 4
…………
A3 - A2 = 2 * 3 - 4
A2 - A1 = 2 * 2 - 4
把上面这些等式左、右两边分别相加，得到：
An - A1 = 2 * [n + (n-1) + (n-2) + …… + 3 + 2] - (n-1) * 4
= 2 * [1 + 2 + 3 + …… + (n-1) + n - 1] - 4(n-1)
= 2 * n(n+1)/2 - 2 - 4n + 4
= n(n+1) - 4n + 2
= n² + n - 4n + 2
= n² - 3n + 2
那么：
An = n² - 3n + 2 + A1 = n² - 3n + 2 = (n-1)(n-2)
验证，当 n = 1 时，A1 = (n-1)(n-2) = 0
An+1 = (n+1 -1)(n+1 - 2) = n(n-1)
An+1 - An = n(n-1) - (n-1)(n-2) = (n-1) * [n - (n-2)] = 2(n-1)
所以，An = (n-1)(n-2) 是数列 {An} 的通项公式。