If ${ }^{20} \mathrm{C}_{1}+\left(2^{2}\right){ }^{20} \mathrm{C}_{2}+\left(3^{2}\right){ }^{20} \mathrm{C}_{3}+\ldots \ldots \ldots .+\left(20^{2}\right)^{20} \mathrm{C}_{20}=\mathrm{A}\left(2^{\beta}\right)$, then the ordered pair $(A, \beta)$ is equal to :
Correct Option: , 2
Given, ${ }^{20} C_{1}+2^{2} \cdot{ }^{20} C_{2}+3^{2} \cdot{ }^{20} C_{3}+\ldots+20^{2} \cdot{ }^{20} C_{20}=A\left(2^{\beta}\right)$
Taking L.H.S.,
$=\sum_{r=1}^{20} r^{2}{ }^{20} C_{r}=20 \sum_{r=1}^{20} r \cdot{ }^{19} C_{r-1}$
$=20\left[\sum_{r=1}^{20}(r-1){ }^{19} C_{r-1}+\sum_{r=1}^{20}{ }^{19} C_{r-1}\right]$
$=20\left[19 \sum_{r=2}^{20}{ }^{18} C_{r-2}+2^{19}\right]=20\left[19 \cdot 2^{18}+2^{19}\right]$
$=420 \times 2^{18}$
Now, compare it with R.H.S., $A=420$ and $\beta=18$
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