1. Write the general term

The given sum is

$$(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17}+\cdots+20(21)^{19}.$$

Its general term is

$$(r+1)(21)^r(20)^{19-r}, \quad r=0,1,2,\dots,19.$$

So

$$S=\sum_{r=0}^{19}(r+1)(21)^r(20)^{19-r}.$$

We are given

$$S=k(20)^{19}.$$

Therefore,

$$k=\sum_{r=0}^{19}(r+1)\left(\frac{21}{20}\right)^r.$$

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2. Use a standard identity

We use

$$\sum_{r=0}^{n}(r+1)x^r=\frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(1-x)^2}, \quad x\ne 1.$$

Here, $n=19$ and $x=\frac{21}{20}$.

Thus,

$$k=\frac{1-21\left(\frac{21}{20}\right)^{20}+20\left(\frac{21}{20}\right)^{21}}{\left(1-\frac{21}{20}\right)^2}.$$

Now,

$$1-\frac{21}{20}=-\frac{1}{20}$$

so

$$\left(1-\frac{21}{20}\right)^2=\frac{1}{400}.$$

Hence,

$$k=400\left[1-21\left(\frac{21}{20}\right)^{20}+20\left(\frac{21}{20}\right)^{21}\right].$$

Factor $\left(\frac{21}{20}\right)^{20}$ from the last two terms:

$$-21\left(\frac{21}{20}\right)^{20}+20\left(\frac{21}{20}\right)^{21} =\left(\frac{21}{20}\right)^{20}\left(-21+20\cdot \frac{21}{20}\right)=0.$$

So only $1$ remains:

$$k=400.$$

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3. Alternative quick observation

Notice that

$$\sum_{r=0}^{19}(r+1)a^rb^{19-r}$$

is the expansion-related sum obtained from

$$\frac{d}{da}\left(\sum_{r=0}^{20}a^{r}b^{20-r}\right)$$

or directly from the finite-series identity above. Substituting $a=21$, $b=20$ again gives the same result:

$$k=400.$$

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4. Final answer

$$\boxed{400}$$

The derived answer matches the stored correct answer.