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计算:
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+33)=?
人气:223 ℃ 时间:2020-10-02 01:42:05
解答
原式=1+1/((1+2)*2/2)+1/((1+3)*3/2)+...+1/((1+33)*33/2)
=2*(1/(1*2)+1/(2*3)+1/(3*4)...+1/(33*34))
=2*((1-1/2)+(1/2-1/3)+(1/3-1/4)...+(1/33-1/34))
=2*(1-1/34)
=33/17
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