If $1\times 2020\times 2\times 2019\times 3\times 2018+...........2020\times 1=2020\times 2021\times \mathrm{k}$ , then the value of $\frac{\mathrm{k}}{100}$ is equal to

$\begin{array}{c}{\mathrm{T}}_{\mathrm{r}}=\mathrm{r}(2021-\mathrm{r})\\ \mathrm{S}=\sum _{\mathrm{r}=1}^{2020} {\mathrm{T}}_{\mathrm{r}}=2021\sum _{\mathrm{r}=1}^{2020} \mathrm{r}-\sum _{\mathrm{r}=1}^{2020} {\mathrm{r}}^2\\ =2021\times \frac{2020\times 2021}{2}-\frac{2020\times 2021\times 4041}{6}\\ =\frac{2020\times 2021}{2}\left[2021-\frac{4041}{3}\right]\\ =\frac{2020\times 2021}{2}\left[\frac{6063-4041}{3}\right]=\frac{2020\times 2021\times 2022}{6}\\ =2020\times 2021\times 337\end{array}$

It is given that S = 2020 x 2021 x k

Therefore, $\mathrm{k}=337\Rightarrow \frac{\mathrm{k}}{100}=\frac{337}{100}=3.37$