Singapore Mathematical Olympiad (SMO)

Let $x_k = n^{2n} \left( 1 + \frac{1}{n^2} \right)^k$ and $y_k = n^{2n} + (k+1)n^{2n-2}$, $k = 1, \dots, n$. Then
$x_1 < y_1 < x_2 < y_2 < \dots < x_n < y_n$.

By the binomial theorem, we have for $k \ge 2$ and $a \le 1/k^2$,
$$\begin{align*} (1+a)^k &= 1+ka+a \left( \frac{ak(k-1)}{2!} + \frac{a^2k(k-1)(k-2)}{3!} + \dots + \frac{a^{k-1}k!}{k!} \right) \\ &\le 1+ka+a \left( \frac{k(k-1)}{k^2} + \frac{1}{2!} + \frac{k(k-1)(k-2)}{k^4} \frac{1}{3!} + \dots + \frac{k!}{k^2k^{k-2}k!} \right) \\ &\le 1+ka+a \left( \frac{1}{2!} + \dots + \frac{1}{k!} \right) < 1+ka+a \left( \frac{1}{1 \cdot 2} + \dots + \frac{1}{(k-1)!} \right) \\ &\le 1+ka+a \left( 1-\frac{1}{k} \right) < 1+(k+1)a. \end{align*}$$
Let $X_k = \left( 1 + \frac{k}{n^2} \right)^k$, $k = 1, \dots, n$. Then, for $2 \le k \le n$, $\frac{1}{n^2} \le \frac{1}{k^2}$. Therefore
$$1 + \frac{k}{n^2} < X_k < 1 + \frac{k+1}{n^2}.$$
Multiplying throughout by $n^{2n}$, we have
$$n^{2n} + k n^{2n-2} < n^{2n} \left( 1 + \frac{1}{n^2} \right)^k < n^{2n} + (k+1)n^{2n-2}.$$