Binomial Theorem · Question 32

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

Answer

5

Solution

(1 + i 3)12 = k=01212Ck (i 3)k

For odd k = 2r - 1: (i 3)2r-1 = i 3 (i 3)2r-2 = i 3 (-3)r-1

Im (1 + i 3)12 = 3 r=16(-3)r-1 12C2r-1

But 1 + i 3 = 2 ( 3 + i 3 )

(1 + i 3)12 = 212( 4 + i 4 ) = 4096, whose imaginary part is 0

r=16(-3)r-1 12C2r-1 = 0 = 1

Line: x - 3y + 1 = 0

Distance of (12, 3) = |12 - 3 + 1| 1 + 3 = 102 = 5

Answer: 5