If $$^{20}C_1 + (2^2)\,^{20}C_2 + (3^2)\,^{20}C_3 + \ldots + (20^2)\,^{20}C_{20} = A(2^\beta)$$, then the ordered pair $$(A, \beta)$$ is equal to

We begin with the given summation

$$S \;=\;\sum_{k=1}^{20} k^{2}\,{}^{20}C_{k}.$$

To evaluate this, we recall the binomial expansion formula

$$\bigl(1+x\bigr)^{n} \;=\;\sum_{k=0}^{n} {}^{\,n}C_{k}\,x^{k}.$$

From this, two standard derivative identities follow.

1. Differentiating once with respect to $$x$$ gives

$$n(1+x)^{\,n-1} \;=\;\sum_{k=0}^{n} k\,{}^{n}C_{k}\,x^{\,k-1}.$$

Multiplying both sides by $$x$$ we get the useful form

$$\sum_{k=0}^{n} k\,{}^{n}C_{k}\,x^{\,k} \;=\; n\,x\,(1+x)^{\,n-1}.$$

2. Differentiating the identity in step 1 once more and again multiplying by $$x$$ yields

$$\sum_{k=0}^{n} k(k-1)\,{}^{n}C_{k}\,x^{\,k} \;=\; n(n-1)\,x^{2}\,(1+x)^{\,n-2}.$$

Our goal involves $$k^{2}\,{}^{20}C_{k}$$, which we split as

$$k^{2} = k(k-1) + k.$$

Therefore

$$\sum_{k=0}^{20} k^{2}\,{}^{20}C_{k} \;=\;\sum_{k=0}^{20} k(k-1)\,{}^{20}C_{k} \;+\;\sum_{k=0}^{20} k\,{}^{20}C_{k}.$$

We now evaluate each part at $$x=1$$, because setting $$x=1$$ in the derived identities converts every factor $$(1+x)$$ into $$2$$ and removes any residual $$x$$ powers.

First part:

Using the second identity with $$n=20$$ and $$x=1$$, we have

$$\sum_{k=0}^{20} k(k-1)\,{}^{20}C_{k} \;=\; 20\cdot19\cdot1^{2}\,(1+1)^{18} \;=\;20\cdot19\cdot2^{18}.$$

Second part:

Using the first identity with $$n=20$$ and $$x=1$$, we have

$$\sum_{k=0}^{20} k\,{}^{20}C_{k} \;=\;20\cdot1\,(1+1)^{19} \;=\;20\cdot2^{19}.$$

Adding the two evaluated sums gives

$$\sum_{k=0}^{20} k^{2}\,{}^{20}C_{k} \;=\; \bigl(20\cdot19\cdot2^{18}\bigr) \;+\;\bigl(20\cdot2^{19}\bigr).$$

We factor the common powers of $$2$$:

$$20\cdot19\cdot2^{18} + 20\cdot2^{19} \;=\;20\cdot2^{18}\bigl(19 + 2\bigr) \;=\;20\cdot2^{18}\cdot21.$$

Multiplying the integers,

$$20\times21 = 420,$$

so

$$S \;=\;420\,2^{18}.$$

The general form requested in the question is $$A(2^{\beta})$$. By direct comparison, we identify

$$A = 420,\quad \beta = 18.$$

Hence, the correct answer is Option B.