Binomial Theorem – Summation & Distance of Point – JEE Main 28 Jan 2025 Shift 1

Solution:

Let the summation part be $S$.
$$ S = \sum_{r=1}^{6} {}^{12}C_{2r-1} (-3)^{r-1} $$
Let $x = \sqrt{3}i$. Then $x^2 = 3i^2 = -3$.
Substituting $-3$ with $x^2$, the term $(-3)^{r-1}$ becomes $(x^2)^{r-1} = x^{2r-2}$.
$$ S = \sum_{r=1}^{6} {}^{12}C_{2r-1} \frac{x^{2r-1}}{x} $$
$$ S = \frac{1}{x} \left( {}^{12}C_1 x + {}^{12}C_3 x^3 + {}^{12}C_5 x^5 + \dots + {}^{12}C_{11} x^{11} \right) $$
The term in the bracket is the sum of odd terms in the binomial expansion of $(1+x)^{12}$.
Using the formula for the sum of odd terms:
$$ \text{Sum} = \frac{(1+x)^{12} – (1-x)^{12}}{2} $$
Therefore,
$$ S = \frac{1}{x} \left[ \frac{(1+x)^{12} – (1-x)^{12}}{2} \right] $$
Substitute $x = \sqrt{3}i$:
$$ 1+x = 1 + \sqrt{3}i = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2e^{i\pi/3} $$
$$ 1-x = 1 – \sqrt{3}i = 2\left(\frac{1}{2} – i\frac{\sqrt{3}}{2}\right) = 2e^{-i\pi/3} $$
Now calculate the powers:
$$ (1+x)^{12} = (2e^{i\pi/3})^{12} = 2^{12} e^{i4\pi} = 2^{12}(1) = 4096 $$
$$ (1-x)^{12} = (2e^{-i\pi/3})^{12} = 2^{12} e^{-i4\pi} = 2^{12}(1) = 4096 $$
Substitute these back into the expression for $S$:
$$ S = \frac{1}{\sqrt{3}i} \left[ \frac{4096 – 4096}{2} \right] = 0 $$
So, $\alpha = 1 + S = 1 + 0 = 1$.

The equation of the line is $\alpha x – \sqrt{3}y + 1 = 0 \Rightarrow x – \sqrt{3}y + 1 = 0$.
The distance of point $P(12, \sqrt{3})$ from this line is:
$$ d = \frac{|1(12) – \sqrt{3}(\sqrt{3}) + 1|}{\sqrt{1^2 + (-\sqrt{3})^2}} $$
$$ d = \frac{|12 – 3 + 1|}{\sqrt{1 + 3}} $$
$$ d = \frac{10}{\sqrt{4}} = \frac{10}{2} = 5 $$

Ans. 5