Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& \alpha =1+\sum _{\mathrm{r}=1}^6(-1{)}^{\mathrm{r}-1}{}^{12}{\mathrm{C}}_{2\mathrm{r}-1}{3}^{\mathrm{r}-1}\\ & \alpha =1+\sum _{\mathrm{r}=1}^6{}^{12}{\mathrm{C}}_{2\mathrm{r}-1}\frac{(\sqrt{3}\mathrm{i}{)}^{2\mathrm{t}-1}}{\sqrt{3}\mathrm{i}}\mathrm{i}=\text{ iota, let }\sqrt{3}\mathrm{i}=\mathrm{x}\end{array}$ $\begin{array}{rl}\alpha & =1+\frac{1}{\sqrt{3}\mathrm{i}}\left({}^{12}{\mathrm{C}}_1\mathrm{x}+{}^{12}{\mathrm{C}}_3{\mathrm{x}}^3+\dots .{}^{12}{\mathrm{C}}_{11}{\mathrm{x}}^{11}\right)\\ & =1+\frac{1}{\sqrt{3}\mathrm{i}}\left(\frac{(1+\sqrt{3}\mathrm{i}{)}^{12}-(1-\sqrt{3}\mathrm{i}{)}^{12}}{2}\right)\end{array}$ $=1+\frac{1}{\sqrt{3}\mathrm{i}}\left(\frac{{\left(-2{\mathrm{w}}^2\right)}^{12}-(2\mathrm{w}{)}^{12}}{2}\right)=1$ so distance of $(12,\sqrt{3})$ from $x-\sqrt{3}y+1=0$ is $\frac{12-3+1}{2}=5$