find the remainder when
$\displaystyle \sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}$ is divided by $64$
what i try
$$\sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}$$
$$\sum^{2014}_{r=0}(-1)^k\sum^{r}_{k=0}(k+1)(k+2)\binom{2019}{r-k}$$
$$\sum^{0}_{k=0}(k+1)(k+2)\binom{2019}{0-k}-\sum^{1}_{k=0}(k+1)(k+2)\binom{2019}{1-k}+\sum^{2}_{k=0}(k+1)(k+2)\binom{2019}{2-k}+\cdots +\sum^{2014}_{k=0}(k+1)(k+2)\binom{2019}{2014-k}$$
how do i solve it help me please
Let us use the followings :
$$\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}=\frac{-2(r-n)(r-n+1)(r-n+2)}{n(n-1)(n-2)}\binom{n}{r}\tag1$$
$$\sum_{r=0}^{n}r^3\binom nr=2^{n-3}n^2(n+3)\tag2$$ $$\sum_{r=0}^{n}r^2\binom nr=2^{n-2}n(n+1)\tag3$$ $$\sum_{r=0}^{n}r\binom nr=2^{n-1}n\tag4$$ $$\sum_{r=0}^{n}\binom nr=(1+1)^n=2^n\tag5$$ See here, here, here, here for proofs.
Using $(1)$, we get $$\sum^{n-4}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-4-k}=2n-6\tag6$$
$$\sum^{n-3}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-3-k}=2\tag7$$
$$\sum^{n-2}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-2-k}=0\tag8$$
$$\sum^{n-1}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-1-k}=0\tag9$$
$$\sum^{n}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-k}=0\tag{10}$$
Let $n=2019$. Then, our sum can be written as
$$\sum^{n-5}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}$$ $$=\underbrace{\sum^{n}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}}_{A}-\underbrace{\sum^{n}_{r=n-4}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}}_{B}$$
$$=\underbrace{\frac{-2}{n(n-1)(n-2)}\sum^{n}_{r=0}(r-n)(r-n+1)(r-n+2)\binom{n}{r}}_A-\underbrace{\bigg((2n-6)+2\bigg)}_B$$
From $(1)(2)(3)(4)(5)$, $$\sum^{n}_{r=0}(r-n)(r-n+1)(r-n+2)\binom{n}{r}$$ is divisible at least by $2^{2019-3}$. Also, $$n(n-1)(n-2)=2019\times 2018\times 2017$$ is not divisible by $2^2$. So, $A$ is divisible by $2^6=64$.
Conclusion :
$$\sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}=A-B\equiv 0-2\equiv \color{red}{62}\pmod{64}$$
Therefore the total result depends only on the first 4 terms
– Thomas Lesgourgues Feb 07 '19 at 12:24