已知数列an的通项公式为an=1/(n(n+1)(n+2)),求数列an的前n项和Sn_数学解答

已知数列an的通项公式为an=1/(n(n+1)(n+2)),求数列an的前n项和Sn

人气：251 ℃　时间：2019-08-19 02:31:35

解答

an=1/2*[1/n - 2/(n+1) +1/(n+2)]
Sn=1/2{(1/1 -2/2 + 1/3)+(1/2 - 2/3 +1/4)+...+ [1/n - 2/(n+1) +1/(n+2)]}
=1/2[1/1 -1/2 - 1/(n+1) +1/(n+2)]
=1/4-1/[2*(n+1)(n+2)]