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?
On
Hint: $$n(n+1)(n+2)(n+3)$$ $$=n(n+3)\cdot (n+1)(n+2)$$ $$=(n^2+3n)(n^2+3n+2)$$
$$=(n^2+3n-1)(n^2+3n+1)$$ $$=n^4+6n^3+9n^2-1$$
Let $n^4+6n^3+9n^2-1=f(n+1)-f(n)$ where $f(m)=a_0+a_1m+a_2m^2+\cdots$
Clearly, $a_r=0\forall r\ge6$
$n^4+6n^3+9n^2-1$ $=a_1+a_2(2m+1)+a_3(3m^2+3m+1)+a_4(4m^3+6m^2+4m+1)+a_5(5m^4+10m^3+10m^2+5m+1)$
Compare the coefficients of $n^4,n^3,n^2,n,n^0$
$5a_5=1$
$4a_4+10a_5=6$
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 $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}=\frac{2020}5=404.$$