<s>[INST] 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?
 [/INST] Note that:

  
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*$\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 $$(1-3x)^{2/3} = 1 - \frac{2}{3} \cdot \frac{3x}{1!} + \frac{ \frac{2}{3} \cdot \Bigl(\frac{2}{3}-1\Bigr)}{2!} \cdot 9x^{2} - \frac{\frac{2}{3}\cdot \Bigl(\frac{2}{3}-1\Bigr) \Bigl(\frac{2}{3}-2\Bigr)}{3!} \cdot 27x^{3} + \cdots$$
simplify and add the above two equations to get the answer.
You may want to see this thread as well:


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*How can I sum the infinite series  $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$
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