Problem source

\begin{questionparts}
\item
In the binomial expansion of $(2x+1/x^{2})^{6}\;$ for $x\ne0$, show that the term which is independent of $x$ is $240$. Find the term which is independent of $x$ in the binomial expansion of $(ax^3+b/x^{2})^{5n}\,$.
\item Let $\f(x) =(x^6+3x^5)^{1/2}\,$. By considering the expansion  of $(1+3/x)^{1/2}\,$ show that the term which is independent of $x$ in the expansion of $\f(x)$ in powers of $1/x\,$,  for $ \vert x\vert>3\,$, is $27/16\,$. Show that there is no term independent of $x\,$ in the expansion  of $\f(x)$ in powers of $x\,$,  for $ \vert x\vert<3\,$. 
\end{questionparts}

Solution source

\begin{questionparts}
\item The terms will all be of the form $x^{6-3k}$, so we are looking for $\binom{6}{2}2^{4} \cdot 1^2 = 6 \cdot 5 \cdot 8 = 240$

The terms will be $x^{(5n-k)3 -2k} = x^{15n-5k}$, so we want $k = 3n$, $\binom{5n}{3n}a^{5n-3n}b^{5n-2n} = \binom{5n}{3n}a^{2n}b^{3n}$

\item Let $f(x) = (x^6+3x^5)^{1/2}$

If $|x| > 3$, then consider

\begin{align*}
&& f(x) &= x^3(1+3/x)^{1/2} \\
&&&= x^3 \left (1 + \frac12 \frac{3}{x} +\frac{1}{2!} \frac12\cdot \frac{-1}{2} \left ( \frac{3}{x} \right)^2 + \frac{1}{3!} \frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \left (\frac{3}{x} \right)^3 + \cdots  \right) \\
&&&= \cdots \frac{1}{6} \frac{3}{8} 27x^0 + \cdots \\
&&&= \cdots + \frac{27}{16} + \cdots
\end{align*}

If $|x| < 3$ we can consider $f(0) = 0$ and notice that there must be no term independent of $x$

\end{questionparts}