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)
Thanks :)
(1+ax/2)^10 + (1+bx)^10
= (1 + 10(ax/2) + (10)(9)/2 (ax/2)^2 + ... ) + (1 + 10bx + (10)(9)/2 (bx)^2 + ...)
= 2 + (5ax + 10bx) + (45/4) a^2 x^2 + 45b x^2 + ..
= 2 + (5a+10b)x + (45/2 a^2 + 45b^2)x^2 + ...
first term matches your given, but
the 2nd only contains first degree x , not x^2
something is awry here!
taking a guess, we have
45/2 a^2 + 45b^2 = 90
a^2 + 2b^2 = 4
Can't think of any nonzero integers that will do the job.
To find the values of the constants a and b, we can use the binomial theorem to expand the given expression. The binomial theorem allows us to expand expressions of the form (a + b)^n, where n is a positive integer.
The binomial theorem states that for any positive integer n:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
Where C(n, k) is the binomial coefficient, which is given by:
C(n, k) = n! / (k! * (n - k)!)
In our case, we have (1 + ax/2)^10 + (1 + bx)^10, which means we have n = 10, a = a, and b = bx/2.
The first two terms in the expansion of (1 + ax/2)^10 will be the terms involving a^10 and a^9, and the first two terms in the expansion of (1 + bx)^10 will be the terms involving b^10 and b^9.
Let's find these terms and set them equal to 2 and 90x^2, respectively:
From the expansion of (1 + ax/2)^10, we have:
C(10, 0) * a^10 * (x/2)^0 + C(10, 1) * a^9 * (x/2)^1 = 2
Simplifying this equation, we get:
a^10 + 10a^9 * (x/2) = 2
Similarly, from the expansion of (1 + bx)^10, we have:
C(10, 0) * (bx)^0 + C(10, 1) * (bx)^1 = 90x^2
Simplifying this equation, we get:
b^10 + 10b^9 * x = 90x^2
Since a is less than b, we now have a system of equations to solve for a and b:
a^10 + 10a^9 * (x/2) = 2
b^10 + 10b^9 * x = 90x^2
By solving this system of equations, we can find the values of the constants a and b.
Unless the x term had a coeffecient of zero, then of course you wouldn't count it,
so 5a + 10b = 0 ---> a = -2b
and
45/2 a^2 + 45b^2 = 90
(45/2)(4b^2 + 45b^2 = 90
90b^2 + 45b^2 = 90
135b^2 = 90
b^2 = 90/135 = 2/3
b = ±V2/√3
if b = +√2/√2 then a = -2√2/√3
if b = -√2/√3 , then a = 2√2/√3