If alpha = 1 + sum_{r=1}^6 (-3)^{r-1} binom{1 | vedclass

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(B) Given $\alpha = 1 + \sum_{r=1}^6 (-3)^{r-1} \binom{12}{2r-1}$.Consider the expansion of $(1 + \sqrt{3}i)^{12} = \sum_{k=0}^{12} \binom{12}{k} (\sq

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$5$

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(B) Given $\alpha = 1 + \sum_{r=1}^6 (-3)^{r-1} \binom{12}{2r-1}$.Consider the expansion of $(1 + \sqrt{3}i)^{12} = \sum_{k=0}^{12} \binom{12}{k} (\sqrt{3}i)^k$.$(1 + \sqrt{3}i)^{12} = \binom{12}{0} + \binom{12}{1}(\sqrt{3}i) + \binom{12}{2}(\sqrt{3}i)^2 + \binom{12}{3}(\sqrt{3}i)^3 + \dots + \binom{12}{12}(\sqrt{3}i)^{12}$.$(1 - \sqrt{3}i)^{12} = \binom{12}{0} - \binom{12}{1}(\sqrt{3}i) + \binom{12}{2}(\sqrt{3}i)^2 - \binom{12}{3}(\sqrt{3}i)^3 + \dots + \binom{12}{12}(\sqrt{3}i)^{12}$.Subtracting the two:$(1 + \sqrt{3}i)^{12} - (1 - \sqrt{3}i)^{12} = 2 [\binom{12}{1}(\sqrt{3}i) + \binom{12}{3}(\sqrt{3}i)^3 + \dots + \binom{12}{11}(\sqrt{3}i)^{11}]$.Note that $(\sqrt{3}i)^2 = -3$,$(\sqrt{3}i)^4 = 9$,etc.So,$\sum_{r=1}^6 \binom{12}{2r-1} (\sqrt{3}i)^{2r-1} = \frac{(1 + \sqrt{3}i)^{12} - (1 - \sqrt{3}i)^{12}}{2}$.Since $1 + \sqrt{3}i = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i\pi/3}$,then $(1 + \sqrt{3}i)^{12} = 2^{12} e^{i4\pi} = 2^{12}$.Similarly,$(1 - \sqrt{3}i)^{12} = 2^{12} e^{-i4\pi} = 2^{12}$.Thus,the sum is $0$,so $\alpha = 1 + 0 = 1$.The line equation becomes $x - \sqrt{3}y + 1 = 0$.The distance from $(12, \sqrt{3})$ to $x - \sqrt{3}y + 1 = 0$ is $d = \frac{|1(12) - \sqrt{3}(\sqrt{3}) + 1|}{\sqrt{1^2 + (-\sqrt{3})^2}} = \frac{|12 - 3 + 1|}{\sqrt{1 + 3}} = \frac{10}{2} = 5$.

If $\alpha = 1 + \sum_{r=1}^6 (-3)^{r-1} \binom{12}{2r-1}$,then the distance of the point $(12, \sqrt{3})$ from the line $\alpha x - \sqrt{3} y + 1 = 0$ is ..........

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