We need to compute
α = 1 + ∑ r = 1 6 ( − 3 ) r − 1 ( 12 2 r − 1 ) . \alpha=1+\sum_{r=1}^6(-3)^{r-1}\binom{12}{2r-1}. α = 1 + r = 1 ∑ 6 ( − 3 ) r − 1 ( 2 r − 1 12 ) . Notice that the sum involves only odd binomial coefficients.
Use the standard identities from binomial expansion:
( 1 + x ) 12 = ∑ k = 0 12 ( 12 k ) x k , (1+x)^{12}=\sum_{k=0}^{12}\binom{12}{k}x^k, ( 1 + x ) 12 = k = 0 ∑ 12 ( k 12 ) x k , ( 1 − x ) 12 = ∑ k = 0 12 ( 12 k ) ( − x ) k . (1-x)^{12}=\sum_{k=0}^{12}\binom{12}{k}(-x)^k. ( 1 − x ) 12 = k = 0 ∑ 12 ( k 12 ) ( − x ) k . Adding and subtracting these gives separation into even and odd terms.
In particular,
( 1 + x ) 12 − ( 1 − x ) 12 = 2 ∑ k = 1 k odd 12 ( 12 k ) x k . (1+x)^{12}-(1-x)^{12}=2\sum_{\substack{k=1\\ k\text{ odd}}}^{12}\binom{12}{k}x^k. ( 1 + x ) 12 − ( 1 − x ) 12 = 2 k = 1 k odd ∑ 12 ( k 12 ) x k . So,
∑ k = 1 k odd 12 ( 12 k ) x k = ( 1 + x ) 12 − ( 1 − x ) 12 2 . \sum_{\substack{k=1\\ k\text{ odd}}}^{12}\binom{12}{k}x^k=\frac{(1+x)^{12}-(1-x)^{12}}{2}. k = 1 k odd ∑ 12 ( k 12 ) x k = 2 ( 1 + x ) 12 − ( 1 − x ) 12 .
Our sum is
∑ r = 1 6 ( − 3 ) r − 1 ( 12 2 r − 1 ) . \sum_{r=1}^6(-3)^{r-1}\binom{12}{2r-1}. r = 1 ∑ 6 ( − 3 ) r − 1 ( 2 r − 1 12 ) . Let k = 2 r − 1 k=2r-1 k = 2 r − 1 k k k 1 , 3 , 5 , 7 , 9 , 11 1,3,5,7,9,11 1 , 3 , 5 , 7 , 9 , 11
( − 3 ) r − 1 = ( − 3 ) k − 1 2 . (-3)^{r-1}=(-3)^{\frac{k-1}{2}}. ( − 3 ) r − 1 = ( − 3 ) 2 k − 1 . A better substitution is to write the odd-power sum using x = i 3 x=i\sqrt{3} x = i 3
( i 3 ) 2 r − 1 = i 3 ( − 3 ) r − 1 . (i\sqrt{3})^{2r-1}=i\sqrt{3}\,(-3)^{r-1}. ( i 3 ) 2 r − 1 = i 3 ( − 3 ) r − 1 . Hence,
( − 3 ) r − 1 = ( i 3 ) 2 r − 1 i 3 . (-3)^{r-1}=\frac{(i\sqrt{3})^{2r-1}}{i\sqrt{3}}. ( − 3 ) r − 1 = i 3 ( i 3 ) 2 r − 1 . Therefore
∑ r = 1 6 ( − 3 ) r − 1 ( 12 2 r − 1 ) = 1 i 3 ∑ r = 1 6 ( 12 2 r − 1 ) ( i 3 ) 2 r − 1 . \sum_{r=1}^6(-3)^{r-1}\binom{12}{2r-1}
=\frac{1}{i\sqrt{3}}\sum_{r=1}^6\binom{12}{2r-1}(i\sqrt{3})^{2r-1}. r = 1 ∑ 6 ( − 3 ) r − 1 ( 2 r − 1 12 ) = i 3 1 r = 1 ∑ 6 ( 2 r − 1 12 ) ( i 3 ) 2 r − 1 . Using the odd-term identity,
∑ r = 1 6 ( 12 2 r − 1 ) ( i 3 ) 2 r − 1 = ( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 2 . \sum_{r=1}^6\binom{12}{2r-1}(i\sqrt{3})^{2r-1}
=\frac{(1+i\sqrt{3})^{12}-(1-i\sqrt{3})^{12}}{2}. r = 1 ∑ 6 ( 2 r − 1 12 ) ( i 3 ) 2 r − 1 = 2 ( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 . So
∑ r = 1 6 ( − 3 ) r − 1 ( 12 2 r − 1 ) = ( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 2 i 3 . \sum_{r=1}^6(-3)^{r-1}\binom{12}{2r-1}
=\frac{(1+i\sqrt{3})^{12}-(1-i\sqrt{3})^{12}}{2i\sqrt{3}}. r = 1 ∑ 6 ( − 3 ) r − 1 ( 2 r − 1 12 ) = 2 i 3 ( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 . Thus
α = 1 + ( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 2 i 3 . \alpha=1+\frac{(1+i\sqrt{3})^{12}-(1-i\sqrt{3})^{12}}{2i\sqrt{3}}. α = 1 + 2 i 3 ( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 .
Now simplify the complex numbers.
Observe:
1 + i 3 = 2 ( cos π 3 + i sin π 3 ) , 1+i\sqrt{3}=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right), 1 + i 3 = 2 ( cos 3 π + i sin 3 π ) , 1 − i 3 = 2 ( cos ( − π 3 ) + i sin ( − π 3 ) ) . 1-i\sqrt{3}=2\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right). 1 − i 3 = 2 ( cos ( − 3 π ) + i sin ( − 3 π ) ) . Hence,
( 1 + i 3 ) 12 = 2 12 ( cos 4 π + i sin 4 π ) = 2 12 = 4096 , (1+i\sqrt{3})^{12}=2^{12}\left(\cos 4\pi+i\sin 4\pi\right)=2^{12}=4096, ( 1 + i 3 ) 12 = 2 12 ( cos 4 π + i sin 4 π ) = 2 12 = 4096 , ( 1 − i 3 ) 12 = 2 12 ( cos ( − 4 π ) + i sin ( − 4 π ) ) = 4096. (1-i\sqrt{3})^{12}=2^{12}\left(\cos (-4\pi)+i\sin (-4\pi)\right)=4096. ( 1 − i 3 ) 12 = 2 12 ( cos ( − 4 π ) + i sin ( − 4 π ) ) = 4096. Therefore,
( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 = 0. (1+i\sqrt{3})^{12}-(1-i\sqrt{3})^{12}=0. ( 1 + i 3 ) 12 − ( 1 − i 3 ) 12 = 0. So the summation is 0 0 0
α = 1. \alpha=1. α = 1.
The line becomes
α x − 3 y + 1 = 0 ⟹ x − 3 y + 1 = 0. \alpha x-\sqrt{3}y+1=0 \implies x-\sqrt{3}y+1=0. α x − 3 y + 1 = 0 ⟹ x − 3 y + 1 = 0. We need the distance of point ( 12 , 3 ) (12,\sqrt{3}) ( 12 , 3 )
Distance from point ( x 1 , y 1 ) (x_1,y_1) ( x 1 , y 1 ) A x + B y + C = 0 Ax+By+C=0 A x + B y + C = 0
∣ A x 1 + B y 1 + C ∣ A 2 + B 2 . \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}. A 2 + B 2 ∣ A x 1 + B y 1 + C ∣ . Here,
A = 1 , B = − 3 , C = 1 , A=1,\quad B=-\sqrt{3},\quad C=1, A = 1 , B = − 3 , C = 1 , ( x 1 , y 1 ) = ( 12 , 3 ) . (x_1,y_1)=(12,\sqrt{3}). ( x 1 , y 1 ) = ( 12 , 3 ) . Thus,
Distance = ∣ 1 ⋅ 12 + ( − 3 ) ( 3 ) + 1 ∣ 1 2 + ( − 3 ) 2 . \text{Distance}=\frac{|1\cdot 12+(-\sqrt{3})(\sqrt{3})+1|}{\sqrt{1^2+(\! -\sqrt{3})^2}}. Distance = 1 2 + ( − 3 ) 2 ∣1 ⋅ 12 + ( − 3 ) ( 3 ) + 1∣ . Compute numerator:
12 − 3 + 1 = 10. 12-3+1=10. 12 − 3 + 1 = 10. Compute denominator:
1 + 3 = 2. \sqrt{1+3}=2. 1 + 3 = 2. Hence,
Distance = 10 2 = 5. \text{Distance}=\frac{10}{2}=5. Distance = 2 10 = 5.
Final answer:
5 \boxed{5} 5