Binomial theorem fractional

given that x>0, find the expansion of [4+ (9/x^2)]^0.5 in ascending powers of x up to and including term in x^3.


The ans is 3/x + (2/3)x - (2/27)x^3.
Pls show workings thanksss!

The binomial theorem states that
(1 + t)^n = 1 + nt + (n(n-1)/2!) t^2 + ... .

So, (4 + 9/x^2)^(1/2)
= [(9/x^2)(4x^2/9 + 1)]^(1/2)
= (3/x) (1 + 4x^2/9)^(1/2)
= (3/x) [1 + (1/2) (4x^2/9) + ((1/2)(1/2 - 1)/2!) (4x^2/9)^2 + ...],
by the binomial series with n = 1/2, t = 4x^2/9

= (3/x) [1 + 2x^2/9 - 2x^4/81 + ...]
= 3/x + 2x/3 - 2x^3/27 + ...

I hope this helps!