If $\alpha=1+\displaystyle\sum_{r=1}^{6}(-3)^{r-1}\binom{12}{2r-1}$, then the distance of the point $(12,\sqrt{3})$ from the line $\alpha x-\sqrt{3}\,y+1=0$ is \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: $\sum_{r=1}^{6}(-3)^{r-1}\binom{12}{2r-1}$ is the odd-index part of $(1+x)^{12}$ at $x=\sqrt{3}\,i$, equal to $\dfrac{(1+\sqrt{3}i)^{12}-(1-\sqrt{3}i)^{12}}{2\sqrt{3}\,i}.$ Both powers equal $4096$, so the difference is $0$.
$1\pm\sqrt{3}\,i=2e^{\pm i\pi/3}$, so $(1\pm\sqrt{3}\,i)^{12}=2^{12}e^{\pm 4\pi i}=4096.$
$\displaystyle\sum_{r=1}^{6}\binom{12}{2r-1}(\sqrt{3}\,i)^{2r-1}=\dfrac{(1+\sqrt{3}\,i)^{12}-(1-\sqrt{3}\,i)^{12}}{2}=0.$
Dividing by $\sqrt{3}\,i$ on the left gives $\sum_{r=1}^{6}(-3)^{r-1}\binom{12}{2r-1}=0.$ So $\alpha=1.$
Line: $x-\sqrt{3}\,y+1=0.$ Distance from $(12,\sqrt{3})$:
$$d=\frac{|12-\sqrt{3}\cdot\sqrt{3}+1|}{\sqrt{1+3}}=\frac{|12-3+1|}{2}=\frac{10}{2}=5.$$
Correct Answer: 5