Answer
$\prod_{i=2}^{n+1}\frac{i-1}{(i-1)^2+4}\ or\ \prod_{i=2}^{n+1}\frac{i-1}{i^2-2i+5}$
Work Step by Step
1. Let $i=k+1$, we have $k=i-1$ and when $k=n$, we have $i=n+1$
2. We can rewrite the original expression as
$\prod_{i=2}^{n+1}\frac{i-1}{(i-1)^2+4}=\prod_{i=2}^{n+1}\frac{i-1}{i^2-2i+5}$
