Question

急急急，裂项法计算

1.1/1x2x3+1/2x3x4+…+1/n(n+1)(n+2）

2.2/(X+2)(X+4)+2/(X+4)(X+6)+…+2/(X+2008)(X+2010)

Answer 1

1.解.裂项法.
1/[n(n+1)(n+2)]=(1/2){1/[n)n+1)]-1/[(n+1)(n+2)]}
=(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]

2.解：
原式
=1/2×[1/x-1/(x+2)]+1/2×[1/(x+2)-1/(x+4)]+......+1/2×[1/(x+2006)-1/(x+2008)]
=1/2×[1/x-1/(x+2)+1/(x+2)-1/(x+4)+......+1/(x+2006)-1/(x+2008)]
=1/2×[1/x-1/(x+2008)]
=1/2×[(x+2008)-x]/[x(x+2008)]
=1/2×2008/[x(x+2008)]
=1004/[x(x+2008)]
=1004/(x^2+2008x)

ps:这种方法在数学中叫做‘裂项相消法’。

追问

呃.. 请问下第一题的1/2怎样来的

追答

比如啊
1/1x2x3+1/2x3x4+.....+1/N(N+1)(N+2)=(1/2)*(1/2-1/2*3+1/2*3-1/3*4+...) = (1/2)*(1/2-1/(N+1)(N+2)) < 1/4

Answer 2

由于题目比较复杂，介绍方法如下：
1、分析第n个加数：An=1/[n(n＋1)(n＋2)]=(1/2){1/[n(n＋1)]－1/[(n＋1)(n＋2)]}，则：
和=(1/2){1/[1×2]－1/[(n＋1)(n＋2)]}=……
2、原式=[1/(x＋2)－1/(x＋4)]＋[1/(x＋4)－1/(x＋6)]＋[1/(x＋6)－1/(x＋8)]＋…＋[1/(x＋2008)－1/(x＋2010)]=1/(x＋2)－1/(x＋2010)=…