Question

一道数学题，求解答，要过程

对于任意非零实数a,b 定义运算“*”如下：a*b=ab分之a-b,求2*1+3*2+4*3+...+2012*2011的值
速解，在线等

Answer 1

解：a*b=(a-b)/(ab)=1/b-1/a
=> 2*1=1-1/2
3*2=1/2-1/3
4*3=1/3-1/4
……
2012*2011=1/2011-1/2012
=> 2*1+3*2+4*3+...+2012*2011
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2011-1/2012)
=1-1/2012
=2011/2012

追问

谢谢啦

Answer 2

2*1+3*2+4*3+...+2012*2011
=-1/(1x2)-1/(3x2)....-1/(2012x2011)
=-[1/(1x2)+1/(3x2)...+1/(2012x2011)]
=-(1-1/2+1/2-1/3+1/3...1/2011-1/2012)
=-(1-1/2012)
=-2011/2012

Answer 3

2*1+3*2+4*3+...+2012*2011
=1/(1x2)-1/(3x2)....-1/(2012x2011)
=[1/(1x2)+1/(3x2)...+1/(2012x2011)]
=(1-1/2+1/2-1/3+1/3...1/2011-1/2012)
=(1-1/2012)
=2011/2012

Answer 4

2*1+3*2+4*3+...+2012*2011=(2-1)/2x1+(3-2)/3x2+(4-3)/4x3+...+(2012-2011)/2012x2011=1-1/2+1/2-1/3+1/3-1/4+...+1/2011-1/2012=2011/2012 ok了 ...O(∩_∩)O~

Answer 5

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