Problem 10.4. Determine all non-zero natural numbers $n$ for which the number

$$
S_{n}=1 \cdot C_{2 n}^{1}+2 \cdot C_{2 n}^{2}+3 \cdot C_{2 n}^{3}+\ldots+n \cdot C_{2 n}^{n}
$$

is a perfect square less than 1000000.
Solution. Transform the general term of the sum

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

The sum $S_{n}$ takes the form:

$$
S_{n}=2 n\left(C_{2 n-1}^{0}+C_{2 n-1}^{1}+C_{2 n-1}^{2}+\ldots+C_{2 n-1}^{n-1}\right) .
$$

The sum in parentheses contains $n$ terms - exactly the first half of the sum of all binomial coefficients. Since the binomial coefficients, equally distant from the extreme terms of the expansion, are equal, the sum in parentheses is equal to $2^{2 n-2}$, and then $S_{n}=2 n \cdot 2^{2 n-2}$, i.e., $S_{n}=n \cdot 2^{2 n-1}$.

Since $S_{n} \leq 1000000$, it follows that $n \in\{1,2,3, \ldots, 9\}$. Perfect squares are obtained only for $n=2\left(S_{2}=16\right)$ and for $n=8\left(\text{with} S_{8}=2^{18}=262144\right)$.

## GRADING RUBRIC

Note: Any other correct method of solution is appreciated with full credit.