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 1

5

Answer 2

Given

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

Substitute $r-1 = s$ so that $s$ changes from 0 to 5. Then,

$$\alpha = 1 + \sum_{s=0}^{5} (-3)^s \binom{12}{2s+1}.$$

Notice that

$$(-3)^s = (-1)^s3^s.$$

Now, consider the expansion of

$$(1 + i\sqrt{3})^{12} = \sum_{k=0}^{12} \binom{12}{k}(i\sqrt{3})^k.$$

Separate the sum into even and odd terms:

• For even $k=2r$: $$(i\sqrt{3})^{2r} = (i^2)^r (3)^r = (-1)^r 3^r \quad \text{(real)}.$$
• For odd $k=2r+1$: $$(i\sqrt{3})^{2r+1} = i\sqrt{3}\,(i^2)^r3^r = i\sqrt{3}\,(-1)^r3^r \quad \text{(imaginary)}.$$

So the imaginary part of the expansion is:

$$\sqrt{3} \sum_{r=0}^{5} \binom{12}{2r+1} (-1)^r3^r.$$

But note that

$$1 + i\sqrt{3} = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right),$$

so

$$(1 + i\sqrt{3})^{12} = 2^{12}\left(\cos 4\pi+i\sin 4\pi\right) = 4096.$$

Since the result is real, the imaginary part must equal zero:

$$\sqrt{3} \sum_{r=0}^{5} \binom{12}{2r+1} (-1)^r3^r = 0 \quad\Longrightarrow\quad \sum_{r=0}^{5} \binom{12}{2r+1} (-1)^r3^r=0.$$

This sum is exactly the one in $\alpha$:

$$\alpha = 1 + \sum_{r=0}^{5} \binom{12}{2r+1}(-3)^r = 1 + 0 = 1.$$

Now, the distance $d$ from the point $(12,\, \sqrt{3})$ to the line

$$\alpha x - \sqrt{3}y + 1=0$$

with $\alpha=1$ becomes:

$$d = \frac{|1\cdot12 - \sqrt{3}\cdot\sqrt{3} +1|}{\sqrt{1^2+(-\sqrt{3})^2}} = \frac{|12 - 3 + 1|}{\sqrt{1+3}} = \frac{10}{2} = 5.$$