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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Binomial Theorem

^ If α=1+∑_ r=1 ^6(-3)^ r-1 ^ 12 C_ 2 r-1 , then the distance of...

 { }^{\text{ } } If α=1+∑r=16(−3)r−112\alpha=1+\sum_{r=1}^6(-3)^{r-1} \quad{ }^{12}\mathrm{C}_{2 r-1},thenthedistanceofthepoint, then the distance of the point(12, \sqrt{3})fromthelinefrom the line\alpha x-\sqrt{3} y+1=0is _________.

Correct answer

0

Step-by-step solution

α=1+∑r=16(−1)r−112C2r−13r−1α=1+∑r=1612C2r−1(3i)2t−13ii= iota, let 3i=x\begin{aligned} & \alpha=1+\sum_{\mathrm{r}=1}^6(-1)^{\mathrm{r}-1}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} 3^{\mathrm{r}-1} \\ & \alpha=1+\sum_{\mathrm{r}=1}^6{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} \frac{(\sqrt{3} \mathrm{i})^{2 \mathrm{t}-1}}{\sqrt{3} \mathrm{i}} \quad \mathrm{i}=\text{ iota, let } \sqrt{3} \mathrm{i}=\mathrm{x}\end{aligned}α=1+13i(12C1x+12C3x3+….12C11x11)=1+13i((1+3i)12−(1−3i)122)\begin{aligned} \alpha & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left({ }^{12} \mathrm{C}_1 \mathrm{x}+{ }^{12} \mathrm{C}_3 \mathrm{x}^3+\ldots .{ }^{12} \mathrm{C}_{11} \mathrm{x}^{11}\right) \\ & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{(1+\sqrt{3} \mathrm{i})^{12}-(1-\sqrt{3} \mathrm{i})^{12}}{2}\right)\end{aligned} =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{\left(-2 \mathrm{w}^2\right)^{12}-(2 \mathrm{w})^{12}}{2}\right)=1sodistanceofso distance of(12, \sqrt{3})fromfromx-\sqrt{3} y+1=0isis\frac{12-3+1}{2}=5
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