题目

裂项相消法求和（1）an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题
裂项相消法求和（1）an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题

提问时间：2021-04-04

答案

(1)an=(1/2)[1/(2n+1)-1/(2n+3)],∴a1+a2+……+an=(1/2)[1/3-1/5+1/5-1/7+……+1/（2n+1)-1/(2n+3)]=(1/2)[1/3-1/(2n+3)]=n/(6n+9).先把an裂项,求和时注意消去哪些项,留下哪些项.余者类推.剩下的题目留给您练习....