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??

see here https://en.wikipedia.org/wiki/Binomial_theorem – Dr. Sonnhard Graubner Jan 20 '18 at 15:35 — Dr. Sonnhard Graubner, Jan 20 '18 at 15:35

score 2 · Answer 1 · edited Jul 14 '18 at 01:39

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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.

edited Jul 14 '18 at 01:39

Fawad

2,034

answered Jan 20 '18 at 15:37

Student

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Here you have to keep in mind that this works $|x|<0.5.$ – Student Jan 20 '18 at 15:39
Why is it only valid for $|x|<0.5$. and this expansion is infinitely on, so what about rational square roots whose expansion is not really never ending. So is there isn't any finite expansion or is it controversial or something?? – Sri Jan 21 '18 at 07:49

score 0 · Answer 2 · answered Jan 20 '18 at 15:44

0

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.

answered Jan 20 '18 at 15:44

QuIcKmAtHs

1,451

@Sri what do you think? – QuIcKmAtHs Jan 20 '18 at 15:57
Please see my comment on supermario's post. Thanks – Sri Jan 21 '18 at 07:49

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

2 Answers2