When (1-x)(1+ax)^6 is expanded as far as the term term in x^2, the result is 1+bx^2. Find the possible values of a and b.
= (1 - x)(1 + 6ax + 15a²x² + ...)
= 1 + 6ax + 15a²x² - x - 6ax² + ...
= 1 + (6a - 1) x + (15a² - 6a)x² + ...
So, 6a - 1 = 0
a = 1/6
b = 15a² - 6a = 15/36 - 1 = -21/36
Did all mentally and without calculator... so do check prudently ;)
Originally posted by eagle:= (1 - x)(1 + 6ax + 15a²x² + ...)
= 1 + 6ax + 15a²x² - x - 6ax² + ...
= 1 + (6a - 1) x + (15a² - 6a)x² + ...
So, 6a - 1 = 0
a = 1/6
b = 15a² - 6a = 15/36 - 1 = -21/36
Did all mentally and without calculator... so do check prudently ;)
Eagle, I love you.