Let's determine the following sums (as functions of $n$):

$$
\begin{aligned}
& A=2^{2}+4^{2}+6^{2}+\ldots+(2 n-2)^{2} \\
& B=1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}
\end{aligned}
$$
We start from the assumption that

$$
1^{2}+2^{2}+3^{2}+\ldots+k^{2}=\frac{1}{6} k(k+1)(2 k+1)
$$

The sum $A+B$ can be expressed based on the previously written relation, if, for instance, $k=2 n-1$. Accordingly,

$$
\begin{gathered}
A+B=1^{2}+2^{2}+3^{2}+\ldots+(2 n-2)^{2}+(2 n-1)^{2}=\frac{1}{6}(2 n-1) 2 n(4 n-1)= \\
=\frac{1}{3} n(2 n-1)(4 n-1) \ldots
\end{gathered}
$$

$$
\begin{aligned}
& A-B=\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+\left(6^{2}-5^{2}\right)+\ldots+\left[(2 n-2)^{2}-(2 n-3)^{2}\right]-(2 n-1)^{2}= \\
& =[3+7+11+\ldots+(4 n-5)] * *^{0}-(2 n-1)^{2}=\frac{(4 n-5+3)(n-1)}{2}-(2 n-1)^{2}= \\
& =(2 n-1)(n-1)-(2 n-1)^{2}=(2 n-1)[n-1-(2 n-1)]=-n(2 n-1) . \\
& 2 A=\frac{1}{3} n(2 n-1)(4 n-1)-n(2 n-1)=n(2 n-1)\left[\frac{4 n-1}{3}-1\right]=\frac{4}{3} n(2 n-1)(n-1) \\
& \qquad A=\frac{2}{3} n(n-1)(2 n-1) \\
& 2 B=\frac{1}{3} n(2 n-1)(4 n-1)+n(2 n-1)=n(2 n-1)\left[\frac{4 n-1}{3}+1\right]=\frac{2 n}{3}(2 n-1)(2 n+1) \\
& B=\frac{n}{3}\left(4 n^{2}-1\right) .
\end{aligned}
$$

János Hörcher (Érseki rg. VII. o. Bp. II.).[^0]


[^0]:    ${ }^{0}$ We have a first-order arithmetic progression within the square brackets, where the number of terms is $n-1$. Indeed, the number of terms in $A$ is $n-1$, and the number of terms in the sum $B$ is $n$.