Let (2x^2 + 3x + 4)^{10} = sum_{r=0}^{20} a_r | vedclass

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(D) Given $(2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r$. Replace $x$ with $\frac{2}{x}$ in the identity: $(2(\frac{2}{x})^2 + 3(\frac{2}{x}) + 4)^{

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$8$

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(D) Given $(2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r$. Replace $x$ with $\frac{2}{x}$ in the identity: $(2(\frac{2}{x})^2 + 3(\frac{2}{x}) + 4)^{10} = \sum_{r=0}^{20} a_r (\frac{2}{x})^r$. $(\frac{8 + 6x + 4x^2}{x^2})^{10} = \sum_{r=0}^{20} a_r 2^r x^{-r}$. $\frac{2^{10}(2x^2 + 3x + 4)^{10}}{x^{20}} = \sum_{r=0}^{20} a_r 2^r x^{-r}$. $2^{10} \sum_{r=0}^{20} a_r x^r = \sum_{r=0}^{20} a_r 2^r x^{20-r}$. To find the ratio $\frac{a_7}{a_{13}}$,we compare the coefficients of $x^7$ on both sides. On the $L$.$H$.$S$.,the coefficient of $x^7$ is $2^{10} a_7$. On the $R$.$H$.$S$.,we set $20-r = 7$,which gives $r = 13$. The coefficient is $a_{13} 2^{13}$. Equating them: $2^{10} a_7 = a_{13} 2^{13}$. Therefore,$\frac{a_7}{a_{13}} = \frac{2^{13}}{2^{10}} = 2^3 = 8$.

Let $(2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r$. Then $\frac{a_7}{a_{13}}$ is equal to

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