<s>[INST] Value of $\sum_{r= 0}^{2019}\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{2021}{r-k} $ What is tried was : $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\mbox{Note that inside summation is }\quad\left.
\sum_{k = 0}^{r}\pars{-1}^{k}\pars{k + 1}\pars{k + 2}{2021 \choose r - k}
\right\vert_{\ 2021\ >\ 3} =
\left.\partiald[2]{}{x}\sum_{k = 0}^{r}{2021 \choose r - k}x^{k + 2}
\right\vert_{\ x\ =\ -1}
\end{align}  , now how to simplfiy the summation part which needs to be double differentiated or there is some other method to solve for that summation value too ?
 [/INST] In seeking to evaluate
$$\sum_{r=0}^n \sum_{k=0}^r
(-1)^k (k+1) (k+2) {n+2\choose r-k}$$
we write
$$2 \sum_{r=0}^n [z^r] (1+z)^{n+2}
\sum_{k=0}^r {k+2\choose 2} (-1)^k z^k.$$
The coefficient extractor enforces the range of the inner sum:
$$2 \sum_{r=0}^n [z^r] (1+z)^{n+2}
\sum_{k\ge 0} {k+2\choose 2} (-1)^k z^k
\\ = 2 \sum_{r=0}^n [z^r] (1+z)^{n+2}
\frac{1}{(1+z)^3}
= 2 \sum_{r=0}^n [z^r] (1+z)^{n-1}
\\ = 2 \sum_{r=0}^n {n-1\choose r}
= 2 \times 2^{n-1} = 2^n$$
where we have used $n\ge 1,$ for $n=0$ we find $2[z^0] (1+z)^{-1} = 2.$
Addendum. We may also change the order of summation as asked in the
comments to get
$$2 \sum_{k=0}^n {k+2\choose 2} (-1)^k
\sum_{r=k}^n {n+2\choose r-k}
= 2 \sum_{k=0}^n {k+2\choose 2} (-1)^k
\sum_{r=0}^{n-k} {n+2\choose r}
\\ = 2 \sum_{k=0}^n {k+2\choose 2} (-1)^k
[z^{n-k}] \frac{1}{1-z} (1+z)^{n+2}
\\ = 2 [z^n] \frac{1}{1-z} (1+z)^{n+2}
\sum_{k=0}^n {k+2\choose 2} (-1)^k z^k.$$
We once more have a coefficient extractor enforcing the upper limit
of the sum and we find
$$2 [z^n] \frac{1}{1-z} (1+z)^{n+2}
\sum_{k\ge 0} {k+2\choose 2} (-1)^k z^k
= 2 [z^n] \frac{1}{1-z} (1+z)^{n+2} \frac{1}{(1+z)^3}
\\ = 2 [z^n] \frac{1}{1-z} (1+z)^{n-1}
= 2 \sum_{r=0}^{n-1} {n-1\choose r} = 2 \times 2^{n-1} = 2^n.$$
Here we see that keeping the original order of the two summations was
the better method moreover no new features appeared on changing the
order.
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