Question

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

Hint:

For sums involving alternating series and combinations, calculate each term carefully, and for distance calculations, always verify the line's coefficients and the point coordinates.

Answer 1

Step 1: Calculate the value of \(\alpha\).

First, evaluate the constant \(\alpha\) from the given summation:

\[ \alpha = 1 + \sum_{r=1}^{6} (-3)^{r-1} \binom{12}{2r-1} \]

 

Step 2: Determine the distance to the line.

Apply the point-to-line distance formula:

\[ \text{Distance} = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} \]

For the line \(\alpha x - \sqrt{3}y + 1 = 0\) with \(A = \alpha, B = -\sqrt{3}, C = 1\):

\[ \text{Distance} = \frac{|-329 \cdot 12 - \sqrt{3} \cdot \sqrt{3} + 1|}{\sqrt{(-329)^2 + (-\sqrt{3})^2}} \]

\[ = \frac{|-3948 - 3 + 1|}{\sqrt{108241 + 3}} \]

\[ = \frac{3950}{\sqrt{108244}} \approx 5 \]