3. Given that $\mathrm{f}(x)$ is a polynomial of degree 2012, and that $\mathrm{f}(k)=\frac{2}{k}$ for $k=1,2,3, \cdots, 2013$, find the value of $2014 \times \mathrm{f}(2014)$.
3. Answer: 4
Solution. Let $\mathrm{g}(x)=x f(x)-2$, hence $\mathrm{g}(x)$ is a polynomial of degree 2013. Since $\mathrm{g}(1)=$ $\mathrm{g}(2)=\mathrm{g}(3)=\cdots=\mathrm{g}(2013)=0$, we must have
$$
\mathrm{g}(x)=\lambda(x-1)(x-2)(x-3) \cdots(x-2012)(x-2013)
$$
for some $\lambda$. Also, $g(0)=-2=-\lambda \cdot 2013$ !, we thus have $\lambda=\frac{2}{2013 \text { ! }}$ Hence,
$$
g(2014)=\frac{2}{2013!}(2013!)=2014 \cdot f(2014)-2
$$
concluding that $2014 \cdot \mathrm{f}(2014)=4$