2

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?

Maximilian Janisch
  • 13,536
Culver Kwan
  • 2,785
  • 3
    If you note that $$ \begin{align} &\color{#090}{n(n+1)(n+2)(n+3)}\color{#C00}{(n+4)}-\color{#C00}{(n-1)}\color{#090}{n(n+1)(n+2)(n+3)}\ &=[\color{#C00}{(n+4)}-\color{#C00}{(n-1)}]\color{#090}{n(n+1)(n+2)(n+3)}\ &=5n(n+1)(n+2)(n+3) \end{align} $$ you can apply Telescoping Series. – robjohn Jul 10 '19 at 09:53
  • 1
    Essentially a duplicate of https://math.stackexchange.com/questions/2021231/how-can-i-derive-what-is-1-cdot-2-cdot-3-cdot-4-2-cdot-3-cdot-4-cdot-5-3-cdo. – Martin R Jul 10 '19 at 11:33

2 Answers2

10

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 $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}=\frac{2020}5=404.$$

Maximilian Janisch
  • 13,536
1

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 $n^4,n^3,n^2,n,n^0$

$5a_5=1$

$4a_4+10a_5=6$

lab bhattacharjee
  • 274,582