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1/1x2+1/2x3+1/3x4+1/4x5+……1/8x9+1/9x10=?(简便算...

1/1x2+1/2x3+1/3x4+1/4x5+……1/8x9+1/9x10 =1-1/2+1/2-1/3+1/3-1/4+....+1/8-1/9+1/9-1/10 =1-1/10 =9/10

1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10 =1-1/2+1/2-1/3+1/3-....-1/9+1/9-1/10 =1-1/10 =9/10

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!

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7 =1-1/7 =6/7

因为an=n*(n-1)/n+1 所以每个数可以看成2/3+(1+2/4)+(2+2/5)+(3+2/6)...+(8+2/11)= 1+2+3...+8+2/3+2/4+2/5...+2/11=36+2*(1/3+1/4+1/5+...1/11) 因为后面的那个是一个发散数列,好像没有求和公式,所以只能同分计算,当然可以把简单...

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5******+1/98-1/99+1/99-1/100 =1-1/100 =99/100

一般的,有: (n-1)n(n+1) =n^3-n {n^3}求和公式:Sn=[n(n+1)/2]^2 {n}求和公式:Sn=n(n+1)/2 1x2x3+2x3x4+3x4x5+....+7x8x9 =2^3-2+3^3-3+...+8^3-8 =(2^3+3^3+...+8^3)-(2+3+...+8) =[(8*9/2)^2-1]-8*9/2+1 =1260

=2^2-1^2+3^2-2^2+.....33^2-32^2 =33^2-1

1x2/1+2x3/1+3x4/1+4x5/1+.....2011x2012/1 =1-1/2+1/2-1/3+……+1/2011-1/2012 =1-1/2012 =2012分之2011

先求它们的倒数和,再把结果倒回来。 1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9+9x10+10x11 =10×(10+1)×(10+2)÷3 =10×11×12÷3 =110×12÷3 =1320÷3 =440

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