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1/1X2+1/2X3+1/3X4+1/4X5+...+1/2010X2011

做了将近半个小时没做出来,突然想出来了,该题好像在课本上有,觉得太简单了.将题目化为:1/nx(n+1)=1/n---1/(n+1)带入式子得:中间的全消掉了,最后剩下的是1-1/2011=2010/2011!

1/1x2+1/2x3+1/3x4+1/4x5+1/2009x2010+1/2010x2011 =1-1/2+1/2-1/3+1/3-1/4++1/2009-1/2010+ 1/ 2010-1/2011=1-1/2011=2010/2011 注,有公式1/NN+1=1/N - 1/N+1.

1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)++1/(2009x2010) = (1-1/2) + (1/2-1/3)+ (1/3-1/4) + (1/4 -1/5) + + (1/2009 - 1/2010) = 1-1/2 + 1/2-1/3+ 1/3-1/4 + 1/4 -1/5 + + 1/2009 - 1/2010 = 1+(-1/2 + 1/2)+(-1/3+ 1/3)+(-1/4 + 1/4)+( -1/5 + + 1/2009) - 1/2010 = 1 -1/2010 = 2009/2010

解 1/1x2+1/2x3+1/3x4+1/4x5++1/2009x2010 =(1+1/1)+(1+1/2)+(1+1/3)+(1+1/4)++(1+1/2009) =1x2009+2008/2009 =2009+2008/2009

1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7 = (1- 1/2) + (1/2 - 1/3) + (1/6 - 1/7) = 1 - 1/7 = 6/7

解:1/1X2+1/2X3+1/3X4+.+1/2010X2011=(1-1/2)+(1/2-1/3)+.+(1/2010-1/2011)=1-1/2011=2010/20111/1X2+1/2X3+1/3X4+.+1/[n(n+1)]=(1-1/2)+(1/2-1/3)+.+[(1/n-1/(n+1)]=1-1/(n+1)=n/(n+1)类似:1/[a*(a+n)]=(1/n)*[1/a-1/(a+n)]

1/1*2=1-1/21/2*3=1/2-1/31/9*10=1/9-1/10等式两边相加得/1x2+1/2x3+1/3x4+1/4x5+……+1/9x10=1-1/10=9/10

1/1X2+1/2X3+1/3X4+.+1/2010X2011=(1-1/2)+(1/2-1/3)+.+(1/2010-1/2011)=1-1/2011=2010/20111/1X2+1/2X3+1/3X4+.+1/[n(n+1)]=(1-1/2)+(1/2-1/3)+.+[(1/n-1/(n+1)]=1-1/(n+1)=n/(n+1)类似:1/[a*(a+n)]=(1/n)*[1/a-1/(a+n)]例子:1/56=1/(7*8)=1/7-1/81/12=1/(2*6)=(1/4)*[1/2-1/6].等等这个式子是很有用的以上希望对你有所帮助~

1/1*(1+1)+1/2*(2+1)+1/3*(3+1)+……+1/9*(9+1)= 1+1+1+1+1+1+1+1+1+(1/1+1/2+1/3+……+1/9)=9+(1/1+1/2+1/3+……+1/9)=…… 下面好算了吧

=4/5先通分再算

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