a) [ X + 4Y ]3
Using the binomial expansion formula (a + b)n = C(n,0) * a^n + C(n,1) * a^(n-1) * b + C(n,2) * a^(n-2) * b^2 + ... + C(n,n) * b^n
Expanding [ X + 4Y ]3:
= C(3,0) * X^3 + C(3,1) * X^2 * (4Y) + C(3,2) * X * (4Y)^2 + C(3,3) * (4Y)^3
= 1 * X^3 + 3 * X^2 * 4Y + 3 * X * 16Y^2 + 1 * 64Y^3
= X^3 + 12X^2Y + 48XY^2 + 64Y^3
Therefore, [ X + 4Y ]3 = X^3 + 12X^2Y + 48XY^2 + 64Y^3
b) [ X + 2Y ]99 (up to 3rd term)
Using the binomial expansion formula,
n = 99
a = X
b = 2Y
Expanding [ X + 2Y ]99:
= C(99,0) * X^99 + C(99,1) * X^98 * (2Y) + C(99,2) * X^97 * (2Y)^2 + C(99,3) * X^96 * (2Y)^3
= 1 * X^99 + 99 * X^98 * 2Y + 4851 * X^97 * 4Y^2 + 161700 * X^96 * 8Y^3
= X^99 + 198X^98Y + 19404X^97Y^2 + 1293600X^96Y^3
Therefore, [ X + 2Y ]99 up to the 3rd term is X^99 + 198X^98Y + 19404X^97Y^2.
3. Perform the following expansions
a) [ X + 4Y ]3
b) [ X + 2Y ] 99 (up to 3rd term) [3 marks]
1 answer