write down the first four terms of the binomial expansion of (1-y)^8 in ascending power of y.By putting y=1/2x(1-x) in your expansion, find the value of p and q if [1-1/2x(1-x)]^8=1-4x+px^2+qx^3
2 answers
please solve for me
so plug in the Binomial Theorem, and you get
(1-y)^8 = 1 - C(8,1)y + C(8,2)y^2 - C(8,3)y^3 + ...
= 1 - 8y + 28y^2 - 56y^3 + ...
So now let y = 1/2 x(1-x) and you get
(1-y)^8 = 1 - 8(1/2 x(1-x)) + 28(1/2 x(1-x))^2 - 56(1/2 x(1-x))^3 + ...
= 1 - 4x(1-x) + 7x^2(1-x)^2 - 7x^3(1-x^3) + C(8,4)/2^4 x^4(1-x)^4 - ...
Clearly, only the first four terms will involve powers of x up to 3, so that gives
= 1 - 4x + 11x^2 - 21x^3 + ...
(1-y)^8 = 1 - C(8,1)y + C(8,2)y^2 - C(8,3)y^3 + ...
= 1 - 8y + 28y^2 - 56y^3 + ...
So now let y = 1/2 x(1-x) and you get
(1-y)^8 = 1 - 8(1/2 x(1-x)) + 28(1/2 x(1-x))^2 - 56(1/2 x(1-x))^3 + ...
= 1 - 4x(1-x) + 7x^2(1-x)^2 - 7x^3(1-x^3) + C(8,4)/2^4 x^4(1-x)^4 - ...
Clearly, only the first four terms will involve powers of x up to 3, so that gives
= 1 - 4x + 11x^2 - 21x^3 + ...
2 answers