Let's determine the following sum

$$
\sum_{k=0}^{n}\binom{2 k}{k} \cdot\binom{2 n-2 k}{n-k} \cdot \frac{m}{k+m}
$$

where $m$ is a natural number. We interpret the symbol $\binom{0}{0}$ as 1. (By determining the sum, we mean producing an equivalent expression that does not involve the summation symbol.)
$$
\sum_{k=0}^{n}\binom{2 k}{k} \cdot\binom{2 n-2 k}{n-k} \cdot \frac{m}{k+m}
$$

Let's denote the sum by $S(m, n)$. (Otherwise, the summation is not affected by the parameter $m$.) For the cases $n=1$ and $n=2$, with an undetermined $m$,

$$
\begin{gathered}
S(m, 1)=\binom{0}{0}\binom{2}{1} \frac{m}{m}+\binom{2}{1}\binom{0}{0} \frac{m}{1+m}=\frac{2(2 m+1)}{m+1} \\
S(m, 2)=\binom{4}{2}+\binom{2}{1}^{2} \frac{m}{1+m}+\binom{4}{2} \frac{m}{2+m}=\frac{16 m^{2}+32 m+12}{(m+1)(m+2)}= \\
=\frac{4(2 m+1)(2 m+3)}{(m+1)(m+2)}
\end{gathered}
$$

Given the similar expressions for the cases $n=3$ and $n=4$, we conjecture that for every natural number $n$, $S(m, n)$ is equal to the following expression:

$$
\frac{2^{n}(2 m+1)(2 m+3) \cdot \ldots \cdot(2 m+2 n-1)}{(m+1)(m+2) \cdot \ldots \cdot(m+n)}
$$

We will prove this.

For the cases $n=1$ and $n=2$, the expressions for $S(m, n)$ cannot be simplified by the factor containing $m$, so $S(m, n)$ generally cannot be simplified. On the other hand, the denominator in (2) is precisely the product of the $n$ denominators in the terms of (1), making it suitable as the common denominator of the terms. (No other denominator can appear, since the binomial coefficients in (1) are integers, and the factor $\frac{m}{k+m}=1$ for the term with $k=0$.) If we multiply the conjectured identity by the denominator

$$
N(m)=(m+1)(m+2) \cdot \ldots \cdot(m+n)
$$

which is an $n$-th degree polynomial in $m$:

$$
\begin{gathered}
N(m) \cdot S(m, n)=m \cdot N(m) \sum_{k=0}^{n}\binom{2 k}{k}\binom{2 n-2 k}{n-k} \frac{1}{k+m}= \\
=2^{n}(2 m+1)(2 m+3) \cdot \ldots \cdot(2 m+2 n-1)
\end{gathered}
$$

the left-hand side of (2) is the sum of $(n+1)$ terms, each of which is an $n$-th degree polynomial in $m$, so their sum $F(m)$ is also an $n$-th degree polynomial, because the coefficient of $m^n$ is positive in each term. Similarly, the right-hand side of (3) is also an $n$-th degree polynomial in $m$.

Two polynomials of the same degree are identical if and only if their values coincide at one more (distinct) point than their degree. For the proof, we can choose the $(n+1)$ points arbitrarily, let them be $0, -1, -2, \ldots, -n$, i.e., the negatives of the values that the summation index $k$ takes in (1).

Thus, for each $m = -k$, $F(m) = F(-k)$ consists of a single term, because in all other $n$ terms, the factor $k+m=0$ remains. The remaining term is:

$$
F(-k)=(-k)(-k+1) \cdot \ldots \cdot(-2) \cdot(-1) \cdot 1 \cdot 2 \cdot \ldots \cdot(-k+n)\binom{2 k}{k}\binom{2 n-2 k}{n-k}
$$

noting that this notation only means that among the $(n+1)$ factors starting from $(-k)$ and increasing by one, the 0 is omitted; there are $k$ negative factors at the beginning - specifically, there are no negative factors when $m=k=0$ - and then $(n-k)$ positive factors - specifically, there are no positive factors when $m=-n$, i.e., $k=n$.

Thus, the $n$ factors before the binomial coefficients can be combined into two factorials by factoring out the $(-1)$ factors, or into one factorial when $m=0$ and $m=-n$. The binomial coefficients can also be expressed using factorials, so:

$$
F(-k)=(-1)^{k} k!(n-k)!\frac{(2 k)!}{k!\cdot k!} \cdot \frac{(2 n-2 k)!}{(n-k)!\cdot(n-k)!}
$$

We further simplify the numerators using the identity for the product of even and odd factors of $(2r)!$:

$$
\begin{gathered}
(2 r)!=(2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 r) \cdot(1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 r-1))= \\
=2^{r} \cdot(1 \cdot 2 \cdot 3 \cdot \ldots \cdot r) \cdot(1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 r-1)) \\
\frac{(2 r)!}{r!}=2^{r} \cdot 1 \cdot 3 \cdot \ldots \cdot(2 r-1)
\end{gathered}
$$

Thus,

$$
F(-k)=(-1)^{k} \cdot 2^{n} \cdot 1 \cdot 3 \cdot \ldots \cdot(2 k-1) \cdot 1 \cdot 3 \cdot \ldots \cdot(2 n-2 k-1)
$$

noting again that the first $k$ factors are followed by $(n-k)$ factors, increasing by 2 from 1, and this is true for $k=0$ and $k=n$ as well.

If we multiply the $(-1)^{k}$ factor, i.e., $k$ factors of $(-1)$, into the first $k$ factors and reverse their order, then

$$
F(-k)=2^{n}(-2 k+1)(-2 k+3) \cdot \ldots \cdot(-3)(-1) \cdot 1 \cdot 3 \cdot \ldots \cdot(2 n-2 k-1)
$$

which is precisely the value of $G(m)$ at $m=-k$ [see the right-hand side of (3)].

This completes the proof of the conjectured identity, the sum-free form of (1) is (2), and the problem is solved.

Veres Sándor (Debrecen, Fazekas M. Gymnasium, M. o. t.)

Remarks: 1. The restriction in the problem statement that $m$ is a natural number is not essential: the identity $F(m)=G(m)$ has been proven for all real $m$ using specific values. Similarly, the equality (1) and (2) holds wherever they are defined.

2. The problem essentially required transforming the sum (1) into a product, with the idea that products are generally easier to compute or manipulate further than sums, especially sums of products. However, if we do not even allow the "..." symbol for the sum, we can seemingly eliminate these from (2) by introducing factorials and binomial coefficients. If $m$ is a natural number, the denominator can be written immediately as:

$$
N(m)=\frac{(m+n)!}{m!}
$$

And if we extend (2) by the denominator, and then multiply each of the $n$ new factors in the numerator by 2 and insert them between pairs of the existing factors, we can write this as the ratio of two factorials, and then the entire expression as the ratio of two binomial coefficients:

$$
\begin{gathered}
\frac{(2 m+1)(2 m+2)(2 m+3) \cdot \ldots \cdot(2 m+2 n-1)(2 m+2 n)}{(N(m))^{2}}= \\
=\frac{(2 m+2 n)!m!m!}{(2 m)!(m+n)!(m+n)!}=\frac{\binom{2 m+2 n}{m+n}}{\binom{2 m}{m}}
\end{gathered}
$$

The "..." is hidden in the factorial.

Veres Sándor