观察算式 1\(2x3)=1\2-1\3;1\3x4=1\3-1\4.1\(x+1)(x+2)+1\(x+2)(x+3)+...+1\(x+2007)(x+2008)=2007\(3x+6024)

问题描述:

观察算式 1\(2x3)=1\2-1\3;1\3x4=1\3-1\4.
1\(x+1)(x+2)+1\(x+2)(x+3)+...+1\(x+2007)(x+2008)=2007\(3x+6024)

由算式 1\(2x3)=1\2-1\3;1\3x4=1\3-1\4.得1/n(n+1)=1/n-1/(n-1) 所以 1\(x+1)(x+2)+1\(x+2)(x+3)+...+1\(x+2007)(x+2008)=1/(x+1)-1/(x+2008) =2007/(x+1)(x+2008) 所以(x+1)(x+2008)=(3x+6024) x^2+2...