題目:
\( \frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{403*404} \)
解答:
\( \frac{1}{n*(n+1)} = \frac{1}{n} - \frac{1}{n+1} \)
因此
\( \frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{403*404} \)
= \( \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{403} - \frac{1}{404} \)
= \( 1 - \frac{1}{404} \)
= \( \frac{403}{404} \)
