Fourth term in the expansion of $(1-2x)^{3/2}$

Question

Calculate the fourth term in the expansion of $(1-2x)^{3/2}$

I first tried to use binomial theorem , but of course fractions in combinations cannot be calculated (or is it that I don't know how to). And then is we add $r$ from $0$ to $<n$ by $1$, then there is only 1 term. So how to do it??

Answer 1

You will have to use the general binomial form. With this you will have that $$(1-2x)^{3/2}=1-\dfrac{(3/2)}{1!}(2x)+\dfrac{(3/2)(1/2)}{2!}(2x)^2-\dfrac{(3/2)(1/2)(-1/2)}{3!}(2x)^3+\dfrac{(3/2)(1/2)(-1/2)(-3/2)}{4!}(2x)^4+....$$

Hope this answers your question.

Answer 2

Note the general formula $$(x+y)^n = x^n+\frac{n}{1!}*x^{(n-1)}*y+\frac{n*(n-1)}{2!}*x^{(n-2)}*y^2+\frac{n*(n-1)*(n-2)}{3!}*x^{(n-3)}*y^3 + ...,$$ holds for fractional $n$ too.