The first and the last denominator is important!

Calculus Level 2

Let:

$\displaystyle \frac{1}{50} + \frac{1}{51} + \frac{1}{52} ... + \frac{1}{98} + \frac{1}{99} = \frac{a}{b}$

Does $a$ divisible by $149$ ?

Hints:

Notice that $\displaystyle50+99=149$ , which is as same as $a$ and $149$ is prime.

No Can't do it. Doesn't exist a. Yes

1 solution

Samuel Li
Jan 2, 2015

Consider the fractions pairwise, as $(\frac{1}{50}+\frac{1}{99}) + (\frac{1}{51}+\frac{1}{98})+...$ . Adding pairwise results in $\frac{149}{50*99}+\frac{149}{51*98}+...$ . We can factor out 149, and the expression evaluates to 149 * SomeRandomFraction. Thus, a is divisible by 149.

The first and the last denominator is important!

1 solution

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