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)
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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)
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)
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’.
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+20...
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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)=…
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