If $$1 + (1 - 2^2 \cdot 1) + (1 - 4^2 \cdot 3) + (1 - 6^2 \cdot 5) + \ldots + (1 - 20^2 \cdot 19) = \alpha - 220\beta$$, then an ordered pair $$(\alpha, \beta)$$ is equal to:

We have to evaluate the whole expression

$$1+\bigl(1-2^{2}\cdot1\bigr)+\bigl(1-4^{2}\cdot3\bigr)+\bigl(1-6^{2}\cdot5\bigr)+\ldots+\bigl(1-20^{2}\cdot19\bigr).$$

Let us denote the required sum by $$S.$$ The first term is the isolated $$1.$$ After that, every term is of the form $$1-(2n)^{2}\,(2n-1)$$ where $$n=1,2,3,\ldots,10,$$ because $$2n$$ runs through $$2,4,6,\ldots,20$$ and the corresponding $$2n-1$$ then runs through $$1,3,5,\ldots,19.$$

So we write

$$S \;=\;1+\sum_{n=1}^{10}\Bigl[\,1-(2n)^{2}(2n-1)\Bigr].$$

Next we expand the bracket inside the summation. First, observe

$$(2n)^{2}(2n-1)=4n^{2}\,(2n-1)=8n^{3}-4n^{2}.$$

Hence

$$1-(2n)^{2}(2n-1)=1-\bigl(8n^{3}-4n^{2}\bigr)=1+4n^{2}-8n^{3}.$$

Substituting this back, we have

$$S=1+\sum_{n=1}^{10}\bigl(1+4n^{2}-8n^{3}\bigr).$$

Now we separate the sum term-by-term:

$$S=1+\Bigl[\sum_{n=1}^{10}1+4\sum_{n=1}^{10}n^{2}-8\sum_{n=1}^{10}n^{3}\Bigr].$$

The first summation is simply the count of numbers from 1 to 10:

$$\sum_{n=1}^{10}1=10.$$

For the second summation we recall the standard formula

$$\sum_{n=1}^{m}n^{2}=\frac{m(m+1)(2m+1)}{6}.$$

Putting $$m=10$$ gives

$$\sum_{n=1}^{10}n^{2}=\frac{10\cdot11\cdot21}{6}= \frac{2310}{6}=385.$$

For the third summation we use the well-known identity

$$\sum_{n=1}^{m}n^{3}=\Bigl[\frac{m(m+1)}{2}\Bigr]^{2}.$$

Taking again $$m=10$$ gives

$$\sum_{n=1}^{10}n^{3}= \Bigl[\frac{10\cdot11}{2}\Bigr]^{2}=55^{2}=3025.$$

Substituting every value inside $$S$$ we get

$$S=1+\Bigl[10+4\times385-8\times3025\Bigr].$$

Now we carry out the arithmetic step by step:

$$4\times385=1540,$$

$$8\times3025=24200.$$

Hence the bracket equals

$$10+1540-24200=1550-24200=-22650.$$

Finally, adding the leading $$1$$ outside the bracket,

$$S=1+(-22650)=-22649.$$

The question states that the same number can be written in the form $$\alpha-220\beta.$$ Therefore we must have

$$\alpha-220\beta=-22649.$$

We search among the options for integers $$\alpha,\beta$$ that satisfy this equation. Trying the given pairs:

Option A: $$10-220\cdot97=10-21340=-21330\neq-22649,$$

Option B: $$11-220\cdot103=11-22660=-22649\;(\text{matches}),$$

Option C: $$10-220\cdot103=10-22660=-22650\neq-22649,$$

Option D: $$11-220\cdot97=11-21340=-21329\neq-22649.$$

So the only pair that works is $$\bigl(11,\,103\bigr).$$

Hence, the correct answer is Option 2.