If $$(20)^{19} + 2(21)(20)^{18} + 3(21)^2(20)^{17} + \ldots + 20(21)^{19} = k(20)^{19}$$, then $$k$$ is equal to ______.

We need to find $$k$$ such that $$(20)^{19} + 2(21)(20)^{18} + 3(21)^2(20)^{17} + \cdots + 20(21)^{19} = k(20)^{19}$$.

We can write the given sum in sigma notation as

$$

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

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

$$

It follows that $$k = \displaystyle\sum_{r=0}^{19}(r+1)x^r$$ where $$x = \dfrac{21}{20}$$.

To evaluate this sum, we use the arithmetic-geometric progression formula

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

Substituting $$n = 19$$ and $$x = \dfrac{21}{20}$$ gives

$$1 - x = -\dfrac{1}{20}\quad\text{so}\quad(1-x)^2 = \dfrac{1}{400}\,,$$

and hence

$$

k = \frac{1 - 21\left(\frac{21}{20}\right)^{20} + 20\left(\frac{21}{20}\right)^{21}}{1/400}\,.

$$

It remains to simplify the numerator:

$$

1 - 21\left(\frac{21}{20}\right)^{20} + 20\left(\frac{21}{20}\right)^{21}

= 1 - 21\left(\frac{21}{20}\right)^{20} + 20\cdot\frac{21}{20}\left(\frac{21}{20}\right)^{20}

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

$$

Therefore,

$$k = \dfrac{1}{1/400} = 400\,.$$

The answer is $$400$$.