Expansion concerning the binomial theorem

The question goes:

Expand $(1-2x)^{1/2}-(1-3x)^{2/3}$ as far as the 4th term.

Ans: $x + x^2/2 + 5x^3/6 + 41x^4/24$

How should I do it?

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There should be somewhere in the notes of your course an expansion of $(1+u)^\alpha$ in powers of $u$ around $u=0$ as far as the 4th term. Apply it (twice). –  Did Jul 3 '11 at 9:53
See Binomial series. –  Américo Tavares Jul 3 '11 at 10:14
@Sophia You claim to be in form 4. Where is the binomial theorem covered in the additional/modern mathematics syllabus? –  user38268 Jul 3 '11 at 12:00
Perhaps Sophia is on school holidays. 2 days, 4 answers, no engagement. –  Gerry Myerson Jul 5 '11 at 11:05

Note that:

• $\displaystyle (1-x)^{\frac{p}{q}} = 1 - \frac{p}{q} \cdot \frac{x}{1!} + \frac{ \frac{p}{q} \cdot \Bigl(\frac{p}{q} -1\Bigr)}{2!} \cdot x^{2} - \cdots$

Using this formula you have $$(1-2x)^{1/2} = 1- \frac{1}{2} \cdot \frac{2x}{1!} + \frac{\frac{1}{2} \cdot \Bigl(-\frac{1}{2}\Bigr)}{2!}\cdot 4x^{2} - \frac{ \frac{1}{2} \cdot \Bigl(\frac{1}{2}-1\Bigr) \cdot \Bigl(\frac{1}{2}-2\Bigr)}{3!} \cdot8x^{3} + \cdots$$ and $$(1-3x)^{2/3} = 1 - \frac{2}{3} \cdot \frac{3x}{1!} + \frac{ \frac{2}{3} \cdot \Bigl(\frac{2}{3}-1\Bigr)}{2!} \cdot 9x^{2} - \frac{\frac{2}{3}\cdot \Bigl(\frac{2}{3}-1\Bigr) \Bigl(\frac{2}{3}-2\Bigr)}{3!} \cdot 27x^{3} + \cdots$$

You may want to see this thread as well:

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Chandru: I am curious to know your position on the piece of advice given at meta.math.stackexchange.com/questions/107/…, especially the first, third, fourth and fifth bullets. –  Did Jul 3 '11 at 10:29
@Didier: Well, Didier I guess the student is encountering such type of problems for the first time, so I wouldn't mind in giving him a good answer. –  user9413 Jul 3 '11 at 10:33
@Didier Piau @Chandru Considering the fact that OP is a 16 year old (assuming it's true) and living in malaysia, the typical syllabus there for a 16 year old studying additional mathematics (so called more advanced maths for those in a science stream) is functions,APs, GPs, differentiation, etc. So they are debutants to this sort of thing and the binomial theorem is beyond them (They don't even do it in high school, only in form 6 which is equivalent to NSW's HSC here in Australia) I know this coming from some Malaysians here in Australia. –  user38268 Jul 3 '11 at 11:59
@D Lim, the title of the question contains the term "binomial theorem," so I think it's reasonable to assume Sophia has seen it. And, if the binomial theorem is not permitted, how do you propose to do the problem? –  Gerry Myerson Jul 4 '11 at 0:39
@Gerry Myerson I was only talking about the syllabus in Malaysia as the question has a homework tag to it. I agree that I don't know how to do it apart from crunching numbers with the binomial theorem. By the way we are not even allowed to use the version when the exponent is not an integer in first year, first semester university. So I don't even know how it is a homework question. Perhaps the OP is attending some special class outside school. –  user38268 Jul 4 '11 at 1:03

Substituting $r=1/2$ and $y=-2x$ in the binomial series $$(1 + y)^r = 1 + \frac{r}{{1!}}y + \frac{{r(r - 1)}}{{2!}}y^2 + \frac{{r(r - 1)(r - 2)}}{{3!}}y^3 + \frac{{r(r - 1)(r - 2)(r - 3)}}{{4!}}y^4 + \cdots$$ gives $$(1-2x)^{1/2} = 1 - x - x^2/2 -x^3/2 - (5/8)x^4 + \cdots,$$ while substituting $r=2/3$ and $y=-3x$ gives $$(1-3x)^{2/3} = 1 - 2x - x^2 - (4/3)x^3 - (7/3)x^4 + \cdots.$$ Now take the difference to get the desired result.

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Shai: I am curious to know your position on the piece of advice given at meta.math.stackexchange.com/questions/107/…, especially the first, third, fourth and fifth bullets. –  Did Jul 3 '11 at 10:29

The answers given by Shai Covo and Chandru do the trick and are probably what was intended. Another method is to use Maclaurin series. To find the $k$th term, differentiate $k$ times, set $x$ to zero, and divide by $k$-factorial.

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Hint: Replace $x$ by $-ax$ in the the binomial series

$$\left( 1+x\right) ^{\alpha }=\sum_{k=0}^{\infty }\binom{\alpha }{k}x^{k}$$

and compute for $\alpha =1/2$, $a=2$ and $\alpha =2/3$, $a=3$

$$\sum_{k=0}^{4}\binom{\alpha }{k}\left( -ax\right) ^{k}.$$

The first term of the difference $\left( 1-2x\right) ^{1/2}-\left( 1-3x\right) ^{2/3}$ is $0$. So we have to consider in the last sum $0\le k\le 4$ to evaluate the first $4$ non vanishing terms.

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