求证:1/(1*2)+1/(3*4)+...+1/((2n-1)*2n)=1/(n+1)+1/(n+2)+...+1/2n请给出证明并且解释.

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求证:1/(1*2)+1/(3*4)+...+1/((2n-1)*2n)=1/(n+1)+1/(n+2)+...+1/2n请给出证明并且解释.求证:1/(1*2)+1/(3*4)+...+1/((2n

求证:1/(1*2)+1/(3*4)+...+1/((2n-1)*2n)=1/(n+1)+1/(n+2)+...+1/2n请给出证明并且解释.
求证:1/(1*2)+1/(3*4)+...+1/((2n-1)*2n)=1/(n+1)+1/(n+2)+...+1/2n
请给出证明并且解释.

求证:1/(1*2)+1/(3*4)+...+1/((2n-1)*2n)=1/(n+1)+1/(n+2)+...+1/2n请给出证明并且解释.
1/(1*2)+1/(3*4)+...+1/((2n-1)*2n) =(1-1/2)+(1/3-1/4)+(1/5-1/6)+...+[1/(2n-1)-1/(2n)] =[1+1/3+1/5+..+1/(2n-1)]-[1/2+1/4+1/6+..+1/(2n)] =[1+1/3+1/5+..+1/(2n-1)]-1/2((1+1/2+1/3+..+1/n) --(1) 而 [1+1/2+1/3+..+1/(2n)] =[1/2+1/4+..+1/(2n)]+[1+1/3+1/5+..+1/(2n-1)] (分成偶数项的和与奇数项和) =1/2[1+1/2+..+1/n]+[1+1/3+1/5+..+1/(2n-1)] 上式移项得:[1+1/3+1/5+..+1/(2n-1)]=[1+1/2+1/3+..+1/(2n)]-1/2[1+1/2+..+1/n] --(2) (2)代入(1)得:1/(1*2)+1/(3*4)+...+1/((2n-1)*2n) =[1+1/3+1/5+..+1/(2n-1)]-1/2((1+1/2+1/3+..+1/n) =[1+1/2+1/3+..+1/(2n)]-1/2[1+1/2+..+1/n] -1/2((1+1/2+1/3+..+1/n) =[1+1/2+1/3+..+1/(2n)]-[1+1/2+..+1/n] =1/(n+1)+1/(n+2)+...+1/2n 证毕