数列超简单题1/n(n+2) 前n项和怎么求?

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数列超简单题1/n(n+2) 前n项和怎么求?
数列超简单题
1/n(n+2) 前n项和怎么求?

数列超简单题1/n(n+2) 前n项和怎么求?
因为a1=1/3=1/2(1/1-1/3)
1/n(n+2)=1/2(1/n-1/n+2)
所以前n项和为1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/n-1-1/n+1+1/n-1/n+2)
=1/2(1/1+1/2-1/n+1-1/n+2)
=1/2[3/2-(2n+3)/(n+1)(n+2)]
=3/4-(2n+3)/[2(n+1)(n+2)]

1/n(n+2)=1/2[1/n-1/(n+2)]
会了吧,如果n从1开始,答案是1/2[1-1/(n+2)]

裂项求和
1/n(n+2)=[(1/n) -(1/n+2)]/2
前n项和就为[1-1/3]/2+[(1/2)-(1/4)]/2+[(1/3)-(1/5)]/2+……+[(1/n)-(1/n+2)]/2
=[(1/1)+(1/2)-(1/n+1)-(1/n+2)]/2
=(3/4)+(2n+3)/[2(n+1)(n+2)]