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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \alpha=1+\sum_{\mathrm{r}=1}^6(-1)^{\mathrm{r}-1}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} 3^{\mathrm{r}-1} \ & \alpha=1+\sum_{\mathrm{r}=1}^6{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} \frac{(\sqrt{3} \mathrm{i})^{2 \mathrm{t}-1}}{\sqrt{3} \mathrm{i}} \quad \mathrm{i}=	ext { iota, let } \sqrt{3} \mathrm{i}=\mathrm{x}\end{aligned}$ $\begin{aligned} \alpha & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left({ }^{12} \mathrm{C}_1 \mathrm{x}+{ }^{12} \mathrm{C}_3 \mathrm{x}^3+\ldots .{ }^{12} \mathrm{C}_{11} \mathrm{x}^{11}ight) \ & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{(1+\sqrt{3} \mathrm{i})^{12}-(1-\sqrt{3} \mathrm{i})^{12}}{2}ight)\end{aligned}$ $=1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{\left(-2 \mathrm{w}^2ight)^{12}-(2 \mathrm{w})^{12}}{2}ight)=1$ so distance of $(12, \sqrt{3})$ from $x-\sqrt{3} y+1=0$ is $\frac{12-3+1}{2}=5$

${ }^{\text { } }$ If $\alpha=1+\sum_{r=1}^6(-3)^{r-1} \quad{ }^{12} \mathrm{C}_{2 r-1}$, then the distance…

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