If =1+ _r=1^6(-3)^r-1, ^12C_2r-1, then the distance of the point (12,√3) from the line x-√…

Question

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

Answer: 5

Accepted Answer

$$\alpha = 1+\sum_{r=1}^{6}(-1)^{r-1}{}^{12}C_{2r-1}\,3^{r-1}$$

Let $\sqrt3\,i=x$ (with $i=\sqrt{-1}$). Then $3^{r-1}(-1)^{r-1}=(\sqrt3 i)^{2r-2}\cdot(\text{sign adjustments})$, and

$$\alpha = 1+\frac{1}{\sqrt3\,i}\left({}^{12}C_1x+{}^{12}C_3x^3+\cdots+{}^{12}C_{11}x^{11}\right) = 1+\frac{1}{\sqrt3\,i}\cdot\frac{(1+\sqrt3i)^{12}-(1-\sqrt3i)^{12}}{2}$$

Since $1+\sqrt3i=2\omega^2\cdot(-1)$-type root of unity form (specifically $1+\sqrt3 i = 2e^{i\pi/3}$), evaluating this expression yields $\alpha=1$.

The distance from $(12,\sqrt3)$ to the line $x-\sqrt3y+1=0$ is:

$$d=\frac{|12-3+1|}{\sqrt{1^2+(\sqrt3)^2}}=\frac{10}{2}=5$$

Solutions · 1