If ${ }^{20} \mathrm{C}_{1}+\left(2^{2}\right)^{20} \mathrm{C}_{2}+\left(3^{2}\right)^{20} \mathrm{C}_{3}+\ldots \ldots+\left(20^{2}\right)^{20} \mathrm{C}_{20}$ $=\mathrm{A}\left(2^{\beta}\right)$, then the ordered pair $(A, \beta)$ is equal to:
Correct Option: 1
$(1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots .+{ }^{n} C_{n} x^{n}$
Diff. w.r.t. $x$
$\Rightarrow \mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1}=\mathrm{n}_{1}+\mathrm{n}_{2}(2 \mathrm{x})+\ldots \ldots+\mathrm{n}_{\mathrm{n}} \mathrm{n}(\mathrm{x})^{\mathrm{n}-1}$
Multiply by x both side
$\Rightarrow n x(1+x)^{n-1}=n_{1} x+n C_{2}\left(2 x^{2}\right)+\ldots .+n_{n}\left(n x^{n}\right)$
Diff w.r.t. $\mathrm{x}$
$\Rightarrow \mathrm{n}\left[(1+\mathrm{x})^{\mathrm{n}-1}+(\mathrm{n}-1) \mathrm{x}(1+\mathrm{x})^{\mathrm{n}-2}\right]$
$={ }^{n} C_{1}+{ }^{n} C_{2} 2{ }^{2} x+\ldots .{ }^{n} C_{n}\left(n^{2}\right) x^{n-1}$
Put $\mathrm{x}=1$ and $\mathrm{n}=20$
$\Rightarrow{ }^{20} \mathrm{C}_{1}+2^{2}{ }^{20} \mathrm{C}_{2}+3^{2}{ }^{20} \mathrm{C}_{3}+\ldots .+(20)^{2}{ }^{20} \mathrm{C}_{20}$
$=20 \times 2^{18}[2+19]=420\left(2^{18}\right)=\mathrm{A}\left(2^{\beta}\right)$