<s>[INST] Calculate the product of these $98$ rational numbers $$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) $$
Of course I could do it factor for factor but there has got to be a more efficient way. Does anyone know it?
 [/INST] We have
$$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) =\frac32\times\frac43\cdots\frac{100}{99}=\frac{100}2=50$$
by telescoping.
</s>