## Task B-4.6.

Let $n$ be a natural number. In the expansion of the binomial $\left(\sqrt{x}-\frac{1}{\sqrt[3]{x}}\right)^{n}$, the ratio of the binomial coefficients of the fifth and sixth terms is 1:404. How many terms in this expansion contain a power of $x$ whose exponent is a natural number?
## Solution.

The binomial coefficients of the 5th and 6th terms in the expansion are $\binom{n}{4}$ and $\binom{n}{5}$, respectively. According to the condition of the problem, we have

$$
\frac{\binom{n}{4}}{\binom{n}{5}}=\frac{1}{404}
$$

After expanding the binomial coefficients, we get

$$
\frac{\frac{n \cdot(n-1) \cdot(n-2) \cdot(n-3)}{4 \cdot 3 \cdot 2 \cdot 1}}{\frac{n \cdot(n-1) \cdot(n-2) \cdot(n-3) \cdot(n-4)}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}=\frac{1}{404}
$$

After simplification, we obtain

$$
\frac{5}{n-4}=\frac{1}{404}
$$

which means $n-4=2020$, so $n=2024$.

1 point

Then, in the binomial expansion $\left(\sqrt{x}-\frac{1}{\sqrt[3]{x}}\right)^{2024}$, the general term is

$$
\begin{array}{cc}
\binom{2024}{k} \cdot(\sqrt{x})^{2024-k} \cdot\left(-\frac{1}{\sqrt[3]{x}}\right)^{k}=\binom{2024}{k} \cdot(\sqrt{x})^{2024-k} \cdot(-1)^{k} \cdot\left(\frac{1}{\sqrt[3]{x}}\right)^{k}= & 1 \text{ point} \\
\binom{2024}{k} \cdot(-1)^{k} \cdot x^{\frac{2024-k}{2}} \cdot x^{-\frac{k}{3}}= & 1 \text{ point} \\
\binom{2024}{k} \cdot(-1)^{k} \cdot x^{\frac{6072-5 k}{6}} & 1 \text{ point}
\end{array}
$$

Notice that each term in the expansion contains the power $x^{\frac{6072-5 k}{6}}$. According to the problem, the exponent of this power must be a natural number. Therefore,

$$
\frac{6072-5 k}{6} \in \mathbf{N}, \text{ or } 1012-\frac{5 k}{6} \in \mathbf{N}
$$

This will be satisfied if the given expression is positive and if $k=0$ or a natural multiple of 6.

From $1012-\frac{5 k}{6}>0$, we get $k \leqslant 1214$.

The number of natural multiples of 6 that are less than 1214 is the quotient when 1214 is divided by 6, which is 202. Adding the case $k=0$, the total number of terms is 203.

## Problem B-4.7

The base of a right rectangular pyramid with height $h$ is a rectangle with sides of lengths $a$ and $b$. The length of each lateral edge is $\frac{\sqrt{94}}{2} \text{ cm}$, and the volume of the pyramid is $9 \text{ cm}^3$. If the numbers $a, b$, and $h$ are consecutive terms of an increasing geometric sequence, determine $a$ and $b$.

## Solution.

From the formula for the volume of the pyramid, we have

$$
V=\frac{B h}{3}=\frac{a b h}{3}=9, \quad a b h=27
$$

and because of the property of the geometric sequence, $b^{2}=a h$. 1 point

From the last two equations, we get $b^{3}=27$, or $b=3$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d455efeee432fadf0574g-24.jpg?height=491&width=586&top_left_y=246&top_left_x=655)

For the length of the lateral edge, the height, and half the diagonal of the base, we have

$$
\left(\frac{\sqrt{94}}{2}\right)^{2}=h^{2}+\frac{a^{2}+b^{2}}{4}
$$

or $a^{2}+b^{2}=4\left(\frac{47}{2}-h^{2}\right)$. Since $b=3$ and $b^{2}=a h$, we solve the following system of equations:

$$
a^{2}+4 h^{2}=85, \quad a h=9
$$

This system reduces to the biquadratic equation $a^{4}-85 a^{2}+324=0$, or the equation $\left(a^{2}-4\right)\left(a^{2}-81\right)=0$.

The positive real solutions to this equation are 2 and 9.

Since the given sequence is increasing, the solution to the problem is $a=2 \text{ cm}, b=3 \text{ cm}$.

Note: If someone reduced the system to the biquadratic equation $4 h^{4}-85 h^{2}+81=0$, it follows that $\left(h^{2}-1\right)\left(4 h^{2}-81\right)=0$, so $h_{1}=1 \text{ cm}, h_{2}=4.5 \text{ cm}$. Then, from $a=\frac{9}{h}$ and the monotonicity of the sequence, the final solution is $a=2 \text{ cm}, b=3 \text{ cm}$.