JEE Main 2025 (Online) 28th January Morning ShiftQuestion 1
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 ____.
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$5$
Solution
Concept. The sum is the odd-index part of $(1+i\sqrt3)^{12}$, which is $0$ because that power is real. Formulas used. (1) $1+i\sqrt3=2\left(\cos\frac\pi3+i\sin\frac\pi3\right)$, De Moivre: $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$(2) Distance of $(x_1,y_1)$ from $ax+by+c=0$: $\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$Step 1 - Odd terms of $(1+i\sqrt3)^{12}$. For $k=2r-1$, $$(i\sqrt3)^{2r-1}=i\,(i^2)^{r-1}\,3^{r-1}\sqrt3=i\sqrt3\,(-3)^{r-1}.$$ So the imaginary part of $(1+i\sqrt3)^{12}$ is $\sqrt3\sum_{r=1}^{6}(-3)^{r-1}\,{}^{12}C_{2r-1}$. Step 2 - The power is real. By (1), $(1+i\sqrt3)^{12}=2^{12}(\cos4\pi+i\sin4\pi)=4096$. Its imaginary part is $0$, so the sum is $0$ and $\alpha=1$. Step 3 - Distance. By (2), from $(12,\sqrt3)$ to $x-\sqrt3y+1=0$: $$\frac{|12-3+1|}{\sqrt{1+3}}=\frac{10}{2}=5.$$ Answer: 5