Answer:
a. The coefficient of y^99 in the expansion of (x-y)^99 is (-1)^99 times the coefficient of y^99 in the expansion of (y-x)^99. Using the binomial theorem, we can write:
(y-x)^99 = sum from k=0 to k=99 of (99 choose k) y^k (-x)^(99-k)
The coefficient of y^99 is (99 choose 99) (-x)^0 = 1, so the coefficient of y^99 in the expansion of (x-y)^99 is (-1)^99 times 1, which is -1.
Therefore, the coefficient of y^99 in the expansion of (x-y)^99 is -1.
b. Using the binomial theorem, we can write:
(3x-1)^60 = sum from k=0 to k=60 of (60 choose k) (3x)^k (-1)^(60-k)
The coefficient of x^50 is the coefficient of (3x)^50 times (-1)^(10) in this expansion. Using the binomial theorem again, we can write:
(3x)^50 = (3^50) x^50
The coefficient of x^50 in (3x-1)^60 is therefore:
(60 choose 50) (3^50) (-1)^10 = 8,105,976,473,487,360
Therefore, the coefficient of x^50 in the expansion of (3x-1)^60 is 8,105,976,473,487,360.