If α=1+Σᵣ₌₁⁶(-3)ʳ⁻¹ ¹² C₂ᵣ₋₁, then the distance of the point (12, √3) from the line α x-√3 y+1=0 is ________.

Solution

Maths sir · 4 steps

That messy sum is the imaginary part of (1+i√3)¹², and it vanishes, so α = 1.

1. Spot the pattern. Only odd binomial terms appear, with powers of -3. In the expansion of (1+i√3)¹², the odd terms k = 2r-1 have (i√3)²ʳ⁻¹ = i√3 (-3)ʳ⁻¹. So the imaginary part is √3Σᵣ₌₁⁶(-3)ʳ⁻¹ ¹²C₂ᵣ₋₁.

   (1+i√3)¹² = (2e^(iπ/3))¹² = 2¹²(cos 4π + i sin 4π) = 4096

2. Find alpha. The result 4096 is real, so its imaginary part is 0, and the sum is 0. That leaves α = 1 + 0 = 1, and the line is x - √3y + 1 = 0.
3. Distance formula. For the point (12, √3): d = (|12 - √3·√3 + 1|)/√(1² + (√3)²) = (|12 - 3 + 1|)/√4 = 10/2.
4. So. The distance is 5.

The common slip: 2.5. Forgetting the square root in the denominator, so dividing 10 by 1 + 3 = 4, gives 2.5 instead of 5.

The idea to keep. Alternate-sign sums of only odd (or only even) binomial coefficients with powers of 3 are a signal for (1 ± i√3)ⁿ. Write it in polar form and read off the real or imaginary part.

Answer from the NTA final answer key. Solution by PrepWiser.