What is $\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}$? I encountered the following hard problem in a math olympiad book:

Evaluate $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}.$$

I tried to evaluate $\sum^{k}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}$ where $k=1$ to $10$ and got $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{5}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, \frac{10}{5}$ respectively. How can I prove that the pattern continues?

I will use the following Lemma (that you can prove by induction):
Lemma. For all $N\in\Bbb N:$ $$\sum_{n=1}^N n(n+1)(n+2)(n+3)=\frac15 N(N+1)(N+2)(N+3)(N+4).$$
In our case, we have $$\sum_{n=1}^{2016} n(n+1)(n+2)(n+3) = 2016\cdot2017\cdot2018\cdot2019\cdot2020\cdot\frac15.$$
So $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}=\frac{2020}5=404.$$
