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+
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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)
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
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
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