Let's write down the $n$-th term of the following sequence and the sum of the first $n$ terms: $2,8,16,22,30,36,44,50,58,64, \ldots$
We can only write down the term of a series with a given number at will if we recognize some regularity in the formation of the first ten terms given, and assume that this regularity holds for any term. By forming the difference series of our series (the change of each term relative to the one preceding it), we find that every odd-numbered difference is 6, and every even-numbered difference is 8. Thus, the sum of any two consecutive differences (the change of each term relative to the one two places before it) is 14, according to which the odd and even indexed terms of our series form arithmetic series with a difference of $d=14$:

$$
a_{2 k-1}=2+14(k-1)=14 k-12
$$

(since $2 k-1$ is the $k$-th odd natural number), or, with $2 k-1=n, k=(n+1) / 2$;

$$
a_{n}=2+14 \frac{n-1}{2}=7 n-5, \quad \text { if } n \text { is odd; }
$$

and similarly

$$
a_{2 k}=14 k-6, \text { or } a_{n}=7 n-6, \text { if } n \text { is even. }
$$

We can also write the two expressions for $a_{n}$ in a common form, if on the one hand we use the fact that the larger of two numbers deviates from their arithmetic mean by the same amount "upwards" as the smaller one "downwards", and this deviation is half the span (the absolute value of the difference) of the two numbers, - and on the other hand, the deviation must be taken with alternating signs, which we express with the factor $(-1)^{n}$. Thus, for any index

$$
a_{n}=7 n-\frac{11}{2}-\frac{(-1)^{n}}{2}
$$

(For odd $n$, we need to increase, because $-5>-6$, hence the minus sign in front of $(-1)^{n}$.)

If $n$ is even, $n=2 k$, then in the sum we take $n / 2=k$ numbers from each series, so denoting the sum of the odd and even indexed terms by $S^{\prime}$ and $S^{\prime \prime}$, respectively,

$$
S_{2 k}=S_{k}^{\prime}+S_{k}^{\prime \prime}=\frac{k}{2}\left(2+a_{2 k-1}+8+a_{2 k}\right)=\frac{k}{2}(28 k-8)=14 k^{2}-4 k
$$

with $n=2 k-1$, we get $k$ odd and $k-1$ even indexed terms:

$$
\begin{aligned}
& S_{2 k-1}=S_{k}^{\prime}+S_{k-1}^{\prime \prime}=\frac{k}{2}\left(2+a_{2 k-1}\right)+\frac{k-1}{2}\left(8+a_{2 k-2}\right)= \\
& =\left(7 k^{2}-5 k\right)+\left(7 k^{2}-13 k+6\right)=14 k^{2}-18 k+6
\end{aligned}
$$

(Briefly: $S_{2 k-1}=S_{2 k}-a_{2 k}$ ). Alternatively, with $k=n / 2$, and $k=(n+1) / 2$,

$$
\begin{aligned}
& S_{n}=\frac{7 n^{2}}{2}-2 n \text { if } n \text { is even, } \\
& S_{n}=\frac{7 n^{2}}{2}-2 n+\frac{1}{2} \quad \text { if } n \text { is odd. }
\end{aligned}
$$

Summarized in one formula:

$$
S_{n}=\frac{7 n^{2}-4 n}{2}+\frac{1-(-1)^{n}}{4}
$$

Windisch Andor (Budapest, József A. g. III. o. t.)

Remarks. 1. The statement we made about the difference series can also be phrased as follows: the growth from term to term on average is $(6+8) / 2=7$, so $a_{n}$ is approximately: $a_{n}^{\prime}=a_{1}+7(n-1)$, but we need to subtract 1 from the even indexed terms (nothing from the odd indexed ones). The necessary subtraction can be written with the expression $K_{1}=\left[1+(-1)^{n}\right] / 2$, so $a_{n}=$ $a_{1}+7(n-1)-K_{1}$ and this is identical to (1). Similarly: the number of even terms in the series $1,2,3, \ldots n$ is $K_{2}=\frac{n}{2}-\frac{1-(-1)^{n}}{4}$; and thus

$$
\begin{gathered}
S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)-K_{2} \cdot 1=\frac{n}{2}\left[2 a_{1}+7(n-1)\right]-\frac{n}{2}+\frac{1-(-1)^{n}}{4}= \\
=\frac{n}{2}(7 n-4)+\frac{1-(-1)^{n}}{4}
\end{gathered}
$$

which is identical to (2).

2. We can also arrive at our results if we consider the terms of the series to be the sum of the corresponding terms of the arithmetic series defined by $a_{1}=1.5, d=7$ and the geometric series defined by $a_{1}=0.5, q=-1$.

Katona Mária (Budapest, Szilágyi E. lg. I. o. t.)