化简 1/X×(X+2)+1/(X+2)×(X+4)+…+1/(X+2006)×(X+2008)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/24 15:37:06
化简 1/X×(X+2)+1/(X+2)×(X+4)+…+1/(X+2006)×(X+2008)
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)
这种方法在数学中叫做‘裂项相消法’.
=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)
这种方法在数学中叫做‘裂项相消法’.
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