急,裂项法计算
问题描述：
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)
答
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.原式=1/2×[1/x-1/(x+2)]+1/2×[1/(x+2)-1/(x+4)]+.+1/2×[1/(x+2006)-1/(x+2...呃.. 请问下第一题的1/2怎样来的比如啊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))