查人人中国名人网广电行业发展趋势:求解高一数列求和问题
来源：百度文库 编辑：查人人中国名人网 时间：2024/04/29 17:27:13
求1/(1*2*3)+1/(2*3*4)+……+1/n(n+1)(n+2)的和。
1/[k(k+1)]-1/[(k+1)(k+2)]=2/[k(k+1)(k+2)]
将k=1,2,3,......,n分别代入上式得：
1/(1*2)-1/(2*3)=2/(1*2*3)
1/(2*3)-1/(3*4)=2/(2*3*4)
1/(3*4)-1/(4*5)=2/(3*4*5)
......
1/[n(n+1)]-1/[(n+1)(n+2)]=2/[n(n+1)(n+2)]
将上面式子相加：
1/(1*2)-1/[(n+1)(n+2)]=2[1/(1*2*3)+1/(2*3*4)+……+1/n(n+1)(n+2)]
1/(1*2*3)+1/(2*3*4)+……+1/n(n+1)(n+2)={1/2-1/[(n+1)(n+2)]}/2
1/n(n+1)(n+2)={[1/n(n+1)]-1/[(n+1)(n+2]}/2
全部拆开，中间项抵消了
结果为{1/2-1/[(n+1)(n+2]}/2