Find the constant term in the development of $(2x^2 + \frac{1}{4x})^{99}$ I am struggling with the following problem:
Find the constant term in the development of $(2x^2 + \frac{1}{4x})^{99}$
This is a combinatorics problem where I should find what the constant term is when $x^0$. 
I have made the following developments:
$$\sum\limits_{k=0}^{n=99}  \binom {99} {k} 2^{n-k}x^{2(n-k)} \times (\frac{1}{4x})^k = \sum\limits_{k=0}^{n=99} \binom {99} {k} 2^{99-k}x^{(198-2k)} \times \frac{1}{2^{2k}x^k}$$
This sum terms could be written together like this:
$$\sum\limits_{k=0}^{n=99} \binom {99} {k} \frac{2^{99-k}x^{(198-2k)}}{{2^{2k}x^k}} = \sum\limits_{k=0}^{n=99} \binom {99} {k} 2^{99-3k}x^{(198-3k)}$$
This gives me that $x^0$ when $198-3k = 0$ But this is the wrong answer....
My textbook says that the constant term will be when $3k-99 = 0$ which gives that $k = 33$ and $\binom {99} {33} \times 2^{99-198}$
Where did I make a mistake?
Thank you!

You found that the constant term was the $k=66$ term, which for your setup is correct; you get a coefficient of
$$\binom{99}{66}2^{-99}\;,$$
which is equal to $$\binom{99}{99-66}2^{-99}=\binom{99}{33}2^{-99}\;.$$
Don’t forget that $\dbinom{n}k=\dbinom{n}{n-k}$ always.
If you switch the rôles of the two terms before you apply the binomial theorem, you get the result directly:
$$\begin{align*}
\left(2x^2+\frac1{4x}\right)^{99}&=\sum_{k=0}^{99}\binom{99}k(2x^2)^k\left(\frac1{4x}\right)^{99-k}\\
&=\sum_{k=0}^{99}\binom{99}k2^{k-2(99-k)}x^{2k-(99-k)}\\
&=\sum_{k=0}^{99}\binom{99}k2^{3k-198}x^{3k-99}\;,
\end{align*}$$
so the constant term is the $k=33$ term, whose coefficient is
$$\binom{99}{33}4^{-99}\;.$$
