## Task A-4.2.

Determine all natural numbers $n$ such that some three consecutive coefficients in the expansion of $(1+x)^{n}$ are in the ratio $3: 4: 5$.
## Solution.

Three consecutive coefficients in the expansion of $(1+x)^{n}$ are of the form $\binom{n}{k-1},\binom{n}{k},\binom{n}{k+1}$, where $k$ is a natural number less than $n$.

Since

$$
\begin{aligned}
&\binom{n}{k-1}= \frac{n!}{(k-1)!(n-k+1)!}=\frac{n!}{(k-1)!(n-k)!\cdot(n-k+1)} \\
&\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n!}{(k-1)!(n-k)!\cdot k}
\end{aligned}
$$

after canceling common factors, we get $\frac{3}{4}=\frac{k}{n-k+1}$.

Similarly, we get $\frac{4}{5}=\frac{k+1}{n-k}$.

Rearranging, we obtain the system of equations

$$
\begin{aligned}
& 3 n-7 k=-3 \\
& 4 n-9 k=5
\end{aligned}
$$

whose only solution is $n=62, k=27$.

Thus, the only natural number in the problem is $n=62$.