Question: The first two terms in the expansion of (1+ax/2)^10 + (1+bx)^10 in ascending powers of x., are 2 and 90x^2
Given that a is less than b find the values of the constants a and b . (11 marks)
The Binomial expansions of the two expressions are as follows,
(1+ax/2)^10 = 1 + 10.(ax/2) + 10.9/2.(ax/2)^2 + ...
(1+bx)^10 = 1 + 10.(bx) + 10.9/2.(bx)^2 + ...
Sum = 2 + 10.(ax/2 + bx) + 10.9/2.(a^2x^2/4 + b^2x^2) + ...
Sum = 2 + 10.(a/2 + b)x + 45(a^2/4 + b^2)x^2 + ...
Given information is that Sum = 2 + 90x^2 + ..., hence
10(a/2 + b) = 0 --> a = -2b
45(a^2/4 + b^2) = 90
i.e. a^2/4 + b^2 = 2
4b^2/4 + b^2 = 2 -- using a = -2b
2b^2 = 2
b^2 = 1
b = +/- 1
If b = -1, then a = 2
if b = +1, then a = -2
Since we are given that a < b, then a = -2 and b = 1.
Answer: a = -2, b = 1