12.
จะพบว่า $$P_n(x)=\sum_{k=0}^{n-1} \binom{n-1}{k}(\frac{1}{x-k})$$
พิจารณา $\sum_{k=0}^{n-1} \binom{n-1}{k}(\frac{1}{n-k})$
$=\frac{n}{n}\sum_{k=0}^{n-1} \binom{n-1}{k}(\frac{1}{n-k})$
$=\frac{1}{n}\sum_{k=0}^{n-1} \binom{n}{k}=\frac{2^n-1}{n}$
ดังนั้น$P_{2008}(2008)=\frac{2^{2008}-1}{2008}$
