Binomial Theorum - can you check this answer?Thanks

[Guest]

This is the question:
\(\displaystyle \L
\begin
{\rm Expand }\left( {4 - 3x} \right)^{{\textstyle{1 \over 2}}} {\rm in ascending powers of }x{\rm as far as the term }x^3 ,{\rm stating the values for which the expansion is valid}{\rm .} \\\)


Here is my anser:
\(\displaystyle \L
{\rm Using the formula: }1 + nx + \frac{{n(n - 1)}}{{2!}}x^2 + \frac{{n(n - 1)(n - 2)}}{{3!}}x^3 {\rm gives} \\\)

\(\displaystyle \L
2\left[ {1 - \frac{3}{8}x - \frac{9}{{128}}x^2 - \frac{{27}}{{1024}}x^3 } \right] \\
\\\)

\(\displaystyle \L
{\rm Not sure what values of }x{\rm are valid for the range but I'm guessing:} \\
- \frac{4}{3}{\rm < }\left| x \right|{\rm < }\frac{4}{3} \\\)

Would someone tell me is this anser is right or not?
:?

 

[Unco]

Looks great to me.

Just note that \(\displaystyle \L \mbox{\left|-\frac{3}{4}x\right| < 1 \Rightarrow |x| < \frac{4}{3} \Rightarrow -\frac{4}{3} < x < \frac{4}{3}}\).