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


*

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