If 1^2 cdot ^{20}C_1 + 2^2 cdot ^{20}C_2 + 3^ | vedclass

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(A) We know that $(1+x)^{20} = \sum_{r=0}^{20} {^{20}C_r} x^r = {^{20}C_0} + {^{20}C_1} x + {^{20}C_2} x^2 + \dots + {^{20}C_{20}} x^{20} \dots (i)$Di

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$(420, 18)$

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(A) We know that $(1+x)^{20} = \sum_{r=0}^{20} {^{20}C_r} x^r = {^{20}C_0} + {^{20}C_1} x + {^{20}C_2} x^2 + \dots + {^{20}C_{20}} x^{20} \dots (i)$Differentiating with respect to $x$:$20(1+x)^{19} = ^{20}C_1 + 2 \cdot ^{20}C_2 x + 3 \cdot ^{20}C_3 x^2 + \dots + 20 \cdot ^{20}C_{20} x^{19} \dots (ii)$Multiplying equation $(ii)$ by $x$:$20x(1+x)^{19} = ^{20}C_1 x + 2 \cdot ^{20}C_2 x^2 + 3 \cdot ^{20}C_3 x^3 + \dots + 20 \cdot ^{20}C_{20} x^{20} \dots (iii)$Differentiating equation $(iii)$ with respect to $x$:$20[(1+x)^{19} + 19x(1+x)^{18}] = ^{20}C_1 + 2^2 \cdot ^{20}C_2 x + 3^2 \cdot ^{20}C_3 x^2 + \dots + 20^2 \cdot ^{20}C_{20} x^{19} \dots (iv)$Putting $x=1$ in equation $(iv)$:$20[2^{19} + 19(2^{18})] = 1^2 \cdot ^{20}C_1 + 2^2 \cdot ^{20}C_2 + \dots + 20^2 \cdot ^{20}C_{20}$$= 20 \cdot 2^{18} [2 + 19] = 20 \cdot 21 \cdot 2^{18} = 420 \cdot 2^{18}$Comparing with $A(2^\beta)$,we get $A = 420$ and $\beta = 18$.Thus,the ordered pair is $(420, 18)$.

If $1^2 \cdot ^{20}C_1 + 2^2 \cdot ^{20}C_2 + 3^2 \cdot ^{20}C_3 + \dots + 20^2 \cdot ^{20}C_{20} = A(2^\beta)$,then the ordered pair $(A, \beta)$ is equal to

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