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+

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/20 05:01:28

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)

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+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’.
再问：呃.. 请问下第一题的1/2怎样来的
再答：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

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+ 求和1x2x3+2x3x4+...+n(n+1)(n+2) 1/(1x2x3)+1/(2x3x4)+1/(3x4x5)+.1/(nx(n+1)x(n+2)=? 1x2x3+2x3x4+3x4x5+4x5x6+...+n(n+1)(n+2)= 有理数:1.计算:1/1x2x3+1/2x3x4+1/3x4x5+...+1/98x99x1002.求式子:|x+1|+ 计算1\1x2x3+1\2x3x4+1\3x4x5+``````+1\48x49x50注：(\)为分数线(x)为乘号计算(x^(2n)+x^n+1)(x^(3n)-x^(2n)+1) 计算：(m+2n)/(n-m)+n/(m-n)-2m(n-m)和[(x+2)/(x×x-2x)-(x-1)/(x×x-4 计算(-x)^2n十1.(-x)^n+1 1x2X3+2x3X4+3x4X5+…+7X8X9=? 1x2x3+2x3x4+3x4x5+…+8x9x10 因式分解4x^(n+2)-9x^n+6x^(n-1)-x^(n-2)