化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2007)(x+2008)并求当x=2是代数式的值

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化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2007)(x+2008)并求当x=2是代数式的值化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+

化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2007)(x+2008)并求当x=2是代数式的值
化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2007)(x+2008)
并求当x=2是代数式的值

化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2007)(x+2008)并求当x=2是代数式的值
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-……-1/(x+2007)+1/(x+2007)-1/(x+2008)=1/x-1/(x+2008)=2008/[x(x+2008)]
当x=2时,原式=2008/(2*2010)=502/1005
1/(1*2)+1/(2*3)+1/(3*4)+.+1/(10*11)=1-1/2+1/2-1/3+1/3-1/4+.+1/10-1/11=10/11
这是一个比较重要的式子了!
1/(1*2)+1/(2*3)+1/(3*4)+.+1/{n*(n-1)}=(n-1)/n
类似的还有1/(1*3)+1/(3*5)+1/(5*7)+.+1/[n(n-2)]=1/2{1-1/3+1/3-1/5+.+1/(n-2)-1/n=(n-1)/(2n)
看出规律了吗?分母的两个数之差是多少就提出来个差的倒数
如1/(1*5)+1/(5*9)=1/4(1-1/5+1/5-1/9)=2/9