题目

1/6,1/24,1/60,1/120,------,求通项,并求出前n项和

提问时间：2020-08-11

答案

通项an = 1/[n(n+1)(n+2)]
an = 1/n * [1/(n+1) - 1/(n+2)]
= 1/2 * [1/n - 1/(n+1) - 1/(n+1) + 1/(n+2)]
前n项和Sn = 1/2 * [(1 - 1/2 - 1/2 + 1/3) + (1/2 - 1/3 - 1/3 + 1/4) + (1/3 - 1/4 - 1/4 + 1/5) + ...+ (1/(n-1) - 1/n - 1/n + 1/(n+1)) + (1/n - 1/(n+1) - 1/(n+1) + 1/(n+2))]
= 1/2 * [(1 - 1/2 - 1/(n+1) + 1/(n+2)]
= 1/4 - 1/[2(n+1)(n+2)]