What is the sum of the first $n$ even and the first $n$ odd numbers?
The sum of the first $2,3,4, \ldots, n$ even numbers:

$$
\begin{gathered}
2+4=6=2^{2}+2 \\
2+4+6=12=3^{2}+3 \\
2+4+6+8=20=4^{2}+4 \text { and so } \\
2+4+6+\ldots+2 n=n^{2}+n=n(n+1)
\end{gathered}
$$

The sum of the first $2,3,4, \ldots, n$ odd numbers:

$$
\begin{gathered}
1+3=4=2^{2} \\
1+3+5=9=3^{2} \\
1+3+5+7=16=4^{2} \quad \text { and so } \\
1+3+5+7+\ldots+(2 n-1)=n^{2}
\end{gathered}
$$

(László Vámossy, Budapest.)

Note. Using the sum formula for an arithmetic progression $S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)$, in the first case

$$
S_{n}=\frac{n}{2}(2+2 n)=n(n+1)
$$

and in the second case

$$
S_{n}=\frac{n}{2}(1+2 n-1)=\frac{n}{2} \cdot 2 n=n^{2}
$$