1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+...+1÷(x+2011)(x+2013)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 14:21:03
1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+...+1÷(x+2011)(x+2013)

1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+...+1÷(x+2011)(x+2013)
1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+...+1÷(x+2011)(x+2013)

1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+...+1÷(x+2011)(x+2013)
1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+...+1÷(x+2011)(x+2013)
=1/2*[1/(x+1)-1/(x+3)]+1/2*[1/(x+3)-1/(x+5)]+.+1/2*[1/(x+2011)-1/(x+2013)]
=1/2*[1/(x+1)-1/(x+2013)]
=1/2*[2012/(x+1)(x+2013)]
=1006/[(x+1)(x+2013)]

1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+...+1÷(x+2011)(x+2013)
=(1/2)[1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+...+1/(x+2011)-1/(x+2013)]
=(1/2)[1/(x+1)-1/(x+2013)

写得比较简化,关键步骤看得出应该就懂的

此题很简单,先化简前两项,然后再化简前三项便能找到规律,分母最后化检后的因子为x 1及x 2013,分子为该多项式的项数,根据2n 1=2013可求出

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