Task B-4.2. How many terms in the expansion of the binomial $\left(2 \sqrt{x}-\frac{1}{x^{2}}\right)^{100}, x>0$, have a positive exponent of $x$, and how many of those have a positive even integer exponent of $x$?
## Solution.

The general term of the given binomial is

$$
\binom{100}{k}(2 \sqrt{x})^{100-k}\left(-\frac{1}{x^{2}}\right)^{k}=\binom{100}{k} \cdot 2^{100-k} \cdot(-1)^{k} \cdot\left(x^{\frac{1}{2}}\right)^{100-k} \cdot x^{-2 k}
$$

The power of $x$ is obtained from

$$
\left(x^{\frac{1}{2}}\right)^{100-k} \cdot x^{-2 k}=x^{\frac{100-k}{2}} \cdot x^{-2 k}=x^{\frac{100-5 k}{2}}
$$

The exponent of $x$ is positive if $\frac{100-5 k}{2}>0$, that is,

$$
\begin{aligned}
& 100-5 k>0 \\
& k<20
\end{aligned}
$$

There are 20 such numbers $k$ since $k \in\{0,1,2,3, \ldots, 19\}$.

The exponent of $x$ is positive and even if

$$
\frac{100-5 k}{2}=50-\frac{5}{2} k
$$

is a positive even number. This is possible only if the number $\frac{5}{2} k$ is even, that is, if the number $k$ is divisible by 4 and less than 20. It follows that $k \in\{0,4,8,12,16\}$.