If ^{20}{C_1} + left( {{2^2}} right){,^{20}}{ | vedclass

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Question

$(1+x)^{20}=^{20} C_{0}+^{20} C_{1}+^{20} C_{2} x^{2}+\ldots \ldots+^{20} C_{20} x^{20}.........(i)$

Differential equation w.r.t. $x$

$20(1+x)^{

Vedclass · Accepted Answer

$(420, 18)$

Vedclass · Answer

$(1+x)^{20}=^{20} C_{0}+^{20} C_{1}+^{20} C_{2} x^{2}+\ldots \ldots+^{20} C_{20} x^{20}.........(i)$

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 $x$

$20x 	imes(1+x)^{19}=$

$^{20} \mathrm{C}_{1} \mathrm{x}+2 \cdot^{20} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots \ldots+20^{20} \mathrm{C}_{20} \mathrm{x}^{20}.........(iii)$

Differential equation $(3)$ w.r.t. $x$

$20\left[(1+x)^{19}+19 x(1+x)^{18}ight]=$

$20 	imes ({2^{19}} + {19.2^{18}}) = {1^2}{\,^{20}}{C_1} + {2^2}{\,^{20}}{C_2}+ ...... + {({20^2})^{20}}{C_{20}}$

Put $x=1$ in equation $(iv)$

$20\left(2^{10}+19.2^{18}ight)=1^{200} \mathrm{C}_{1}+2^{2} \mathrm{C}_{2}+\ldots \ldots+\left(20^{2}ight)^{20} \mathrm{C}_{20}$

$=20 	imes 2^{18}(2+19)=20 	imes 21 	imes 2^{18}$

$=420 	imes 2^{18}$

$A=420, \beta=18$

Hence correct Option is $(A)$.

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

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