2008 AMC 10A Problem 5

The general factor is $\frac{4n+4}{4n}$: its numerator $4(n+1)$ is exactly the denominator of the next factor. So in the full product every numerator cancels against the following denominator.

What survives is the very last numerator and the very first denominator:
$$\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdots\frac{2008}{2004} = \frac{2008}{4} = 502.$$

The answer is $\boxed{\textbf{(B)}\ 502}$.

A product of the form $\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdots\frac{a_k}{a_{k-1}}$ telescopes to $\frac{a_k}{a_1}$. Whenever consecutive fractions share a term, look for this cancellation before doing any arithmetic. The $4n$ notation is just $a_n = 4n$, so the product is $a_{502}/a_1$.

Each factor equals $\frac{n+1}{n}$, so the product is $\frac{2}{1}\cdot\frac{3}{2}\cdots\frac{502}{501} = 502$, since $2008 = 4 \cdot 502$.

• Multiplying out the first few fractions numerically, or dividing 2008 by 8 instead of by the first denominator 4.
• Miscounting the number of factors and answering $501$ or $2008$.