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If $^{20}{C_1} + \left( {{2^2}} \right){\,^{20}}{C_3} + \left( {{3^2}} \right){\,^{20}}{C_3} + \left( {{2^2}} \right) + ..... + \left( {{{20}^2}} \right){\,^{20}}{C_{20}} = A\left( {{2^\beta }} \right)$, then the ordered pair $(A, \beta )$ is equal to
$(420, 18)$
$(380, 18)$
$(420, 19)$
$(380, 19)$
Solution
$(1+x)^{20}=^{20} C_{0}+^{20} C_{1}+^{20} C_{2} x^{2}+\ldots \ldots+^{20} C_{20} x^{20}………(i)$
Differential equation w.r.t. $x$
$20(1+x)^{19}=$
$^{20} C_{1} \cdot 1+2.^{20} C_{2} x+\ldots \ldots+20^{20} C_{20} x^{19}………(ii)$
Multiply equation $(2)$ by $x$
$20x \times(1+x)^{19}=$
$^{20} \mathrm{C}_{1} \mathrm{x}+2 \cdot^{20} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots \ldots+20^{20} \mathrm{C}_{20} \mathrm{x}^{20}………(iii)$
Differential equation $(3)$ w.r.t. $x$
$20\left[(1+x)^{19}+19 x(1+x)^{18}\right]=$
$20 \times ({2^{19}} + {19.2^{18}}) = {1^2}{\,^{20}}{C_1} + {2^2}{\,^{20}}{C_2}+ …… + {({20^2})^{20}}{C_{20}}$
Put $x=1$ in equation $(iv)$
$20\left(2^{10}+19.2^{18}\right)=1^{200} \mathrm{C}_{1}+2^{2} \mathrm{C}_{2}+\ldots \ldots+\left(20^{2}\right)^{20} \mathrm{C}_{20}$
$=20 \times 2^{18}(2+19)=20 \times 21 \times 2^{18}$
$=420 \times 2^{18}$
$A=420, \beta=18$
Hence correct Option is $(A)$.
