Consider the sums $$A = \frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{99 \cdot 100}$$ and $$B = \frac{1}{51 \cdot 100} + \frac{1}{52 \cdot 99} + \cdots + \frac{1}{100 \cdot 51}$$. Express $$\frac{A}{B}$$ as an irreducible fraction.

Splitting each term of A as $$\frac{1}{n}-\frac{1}{n+1}$$ and telescoping shows $$A = \frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}$$. For B, factoring out $$\frac{1}{151}$$ from each term (since $$51+100=52+99=\cdots=151$$) and simplifying using partial fractions shows $$B = \frac{2}{151}\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\right)$$. Dividing, the common bracketed sum cancels, giving $$\frac{A}{B} = \frac{151}{2}$$.