If α = 1 + ∑_{r=1}^{6} (-3)^{r-1} ^{12}C_{2r-1} then the distance of the point (12, √3) from the lin

Q: Wait, why do we use complex numbers for a binomial sum?

Great question! The term $(-3)^{r-1}$ is tricky because it's negative. By letting $x = \sqrt{3}i$, we see that $x^2 = -3$. This allows us to transform the sum into a form where we can apply the standard binomial identity for odd-indexed coefficients. It's a powerful tool to turn a seemingly impossible algebraic sum into a clean complex number calculation.

Q: How do I know when to use the $(1+x)^n - (1-x)^n$ identity?

Whenever you see a sum involving only odd-indexed binomial coefficients (like ${}^{n}C_1, {}^{n}C_3, \dots$), this identity is your best friend. If the sum involved even indices, you would use $(1+x)^n + (1-x)^n$. Recognizing the pattern of indices is the first step to choosing the right tool.

Q: Is there a simpler way to calculate $(1+\sqrt{3}i)^{12}$ without polar form?

You could expand it using the binomial theorem, but that would be a nightmare! Polar form (De Moivre's Theorem) is the most elegant and efficient way to handle large powers of complex numbers. It turns a complex expansion into a simple multiplication of the argument by the exponent.

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of binomial coefficients and coordinate geometry.

But fear not! We are going to peel back the layers, one by one, until we find the elegant simplicity hidden at the core.

### Unmasking the Unknown

We start with the expression $\alpha = 1 + \sum_{r=1}^{6} (-3)^{r-1} {}^{12}C_{2r-1}$. This is our mystery.

We have a sum of odd-indexed binomial coefficients, but with that pesky $(-3)^{r-1}$ term. The secret lies in the complex plane.

If we let $x = \sqrt{3}i$, then $x^2 = -3$. Suddenly, the term $(-3)^{r-1}$ becomes $(x^2)^{r-1} = x^{2r-2}$.

Now, our sum looks like this:
$$\sum_{r=1}^{6} x^{2r-2} {}^{12}C_{2r-1}$$

If we factor out $\frac{1}{x}$, we get:
$$\frac{1}{x} \sum_{r=1}^{6} x^{2r-1} {}^{12}C_{2r-1}$$

This is the breakthrough! We have transformed the sum into a standard binomial series involving only odd powers of $x$.

### The Identity of Elegance

Now that we have the series $\sum_{r=1}^{6} x^{2r-1} {}^{12}C_{2r-1}$, we recall the classic JEE identity:
$$\sum_{k 	ext{ odd}} {}^{n}C_k x^k = \frac{(1+x)^n - (1-x)^n}{2}$$

With $n=12$, our sum becomes:
$$\frac{(1+x)^{12} - (1-x)^{12}}{2x}$$

Substituting $x = \sqrt{3}i$, we have:
$$\frac{(1+\sqrt{3}i)^{12} - (1-\sqrt{3}i)^{12}}{2(\sqrt{3}i)}$$

### The Polar Transformation

Do not be afraid of the power of 12. We convert $1+\sqrt{3}i$ into polar form:
$$2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}ight) = 2e^{i\pi/3}$$

Similarly, $1-\sqrt{3}i = 2e^{-i\pi/3}$. Now, raising these to the power of 12 is a breeze:
$$(1+\sqrt{3}i)^{12} = (2e^{i\pi/3})^{12} = 2^{12}e^{i4\pi} = 2^{12}(\cos(4\pi) + i\sin(4\pi)) = 2^{12}$$

Likewise, $(1-\sqrt{3}i)^{12} = 2^{12}$. The numerator is $2^{12} - 2^{12} = 0$.

The entire sum is zero! Thus, $\alpha = 1 + 0 = 1$.

### The Final Distance

With $\alpha = 1$, our line equation simplifies beautifully to $x - \sqrt{3}y + 1 = 0$. We need the distance from the point $(12, \sqrt{3})$ to this line.

Using the perpendicular distance formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$, we substitute $a=1, b=-\sqrt{3}, c=1$ and $(x_1, y_1) = (12, \sqrt{3})$.

The numerator becomes:
$$|1(12) - \sqrt{3}(\sqrt{3}) + 1| = |12 - 3 + 1| = 10$$

The denominator is:
$$\sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$$

Finally, $d = \frac{10}{2} = 5$. We have arrived at our destination: the distance is **5 units**.

Animated Solution for Mathematics - Binomial Theorem: If $α = 1 + \sum_{r = 1}^{6} (- 3)^{r - 1}^{12} C_{2 r - 1}$ then the distance of the point $(12, 3)$ from the line $α x - 3 y + 1 = 0$ is

Enter Numerical Value:

Visualized Solution

Problem Overview

Analyzing $α$

Complex Substitution

Rewriting the Summation

Binomial Identity for Odd Terms

Applying the Identity

Polar Form Conversion

Evaluating the Powers

Calculating $α$

The Line Equation

Visualizing the Distance

Distance Formula

Substituting Values

Simplifying the Expression

Final Answer

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Unmasking the Unknown

The Identity of Elegance

The Polar Transformation

The Final Distance

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α=1+∑r=16​(−3)r−112C2r−1​ then the distance of the point (12,3​) from the line αx−3​y+1=0 is