数列的裂项求和问题a[n]=1/[n(n+2)] 求n项和1/[n(n+2)]=1/n-1/(n+2) 1-1/3+1/2-1/4+1/3-1/5+1/4-1/6 .

a[n]=1/[n(n+2)] 求n项和
1/[n(n+2)]=(1/2){1/n-1/(n+2) }
(1/2){1-1/3+1/2-1/4+1/3-1/5+1/4-1/6 .... }
=(1/2){1+1/2-1/(n+1)-1/(n+2)}
=[3(n+1)(n+2)-2(n+2)-2(n+1)]/4(n+1)(n+2)=(3n+5)n/4(n+1)(n+2)

1-1/3+1/2-1/4+1/3-1/5+1/4-1/6 ....
都不能消项啊.. 怎么做~?
1/3有正负两项，1/4有正负两项。。。。。
怎么不能消？？

Sn=∑(1，n)|(1/(n(n+2)))
=∑(1，n)|(((n+2)-n)/(2n(n+2)))
=(1/2)×∑(1，n)|((1/n)-(1/(n+2)))
=(1/2)×(∑(1，n)|(1/n)-∑(1，n)|(1/(n+2)))
=(1/2)×
[(1+(1/2)
+(1/3)+(1/4)+(1/5)+···+(1/n))
-((1/3)+(1/4)+(1/5)+···+(1/n)+(1/(n+1))+(1/(n+2)))]
=(1/2)×[1+(1/2)-((1/(n+1))+(1/(n+2)))]
=(3n²+5n)/(4×(n+1)(n+2))

a[n]=1/[n(n+2)] =（1/2）(1/n-1/(n+2))
(1/2)(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6 .... )
=(1+1/2-1/(n+1)-1/(n+2))/2

1/[n(n+2)]=[1/n-1/(n+2)]/2
隔2项可以消项
结果为0.5*【1+1/2-1/（n+1）-1/（n+2）】