Let's write down the $n$-th term of the following sequence:

$2,10,16,32,42,66,80,112,130,170, \ldots$
I. Solution: Let the terms of our sequence be: $a_{1}, a_{2}, \ldots, a_{n}$, and the terms of its difference sequence be $d_{1}, d_{2}, \ldots, d_{n-1}$, where $d_{k}=a_{k+1}-a_{k}$. The first nine terms of this are: $8,6,16,10,24,14,32,18,40$. Each even-indexed difference is smaller than the term preceding it, and (excluding $d_{1}$) each odd-indexed one is larger than the term preceding it, because in the so-called second difference sequence $\left(e_{k}=d_{k+1}-d_{k}\right)$, the terms $-2,10,-6,14,-10,18,-14,22$ are of alternating signs.

Relying on this, let's try the odd and even indexed terms of the original sequence separately. Denote these by $a^{\prime}$, $a^{\prime \prime}$, and the terms of the difference sequences by $d^{\prime}, d^{\prime \prime}, e^{\prime}, e^{\prime \prime}$:

$$
\begin{aligned}
& a^{\prime}: \quad 2, \quad 16, \quad 42, \quad 80, \quad 130 \quad a^{\prime \prime}: \quad 10, \quad 32, \quad 66, \quad 112, \quad 170 \\
& d^{\prime}: \quad 14, \quad 26, \quad 38, \quad 50, \quad d^{\prime \prime}: \quad 22, \quad 34, \quad 46, \quad 58 \\
& e^{\prime}: \quad 12, \quad 12, \quad 12 \quad e^{\prime \prime}: \quad 12, \quad 12, \quad 12
\end{aligned}
$$

$e^{\prime}$ and $e^{\prime \prime}$ are constants, so the terms of $d^{\prime}, d^{\prime \prime}$ are arithmetic sequences. Assuming this regularity holds for any term, we have:

$$
\begin{aligned}
& a_{k}^{\prime}=a_{2 k-1}=a_{1}^{\prime}+\left(d_{1}^{\prime}+d_{2}^{\prime}+\ldots+d_{k-1}^{\prime}\right)=a_{1}^{\prime}+\frac{k-1}{2}\left[2 d_{1}^{\prime}+(k-2) e_{1}^{\prime}\right]= \\
& =2+\frac{k-1}{2}[28+12(k-2)]=6 k^{2}-4 k \\
& a_{k}^{\prime \prime}=a_{2 k}=a_{1}^{\prime \prime}+\left(d_{1}^{\prime \prime}+d_{2}^{\prime \prime}+\ldots+d_{k-1}^{\prime \prime}\right)=a_{1}^{\prime \prime}+\frac{k-1}{2}\left[2 d_{1}^{\prime \prime}+(k-2) e_{1}^{\prime \prime}\right]= \\
& =10+\frac{k-1}{2}[44+12(k-2)]=6 k^{2}+4 k
\end{aligned}
$$

For odd and even $n$, respectively, $k=(n+1) / 2-\mathrm{t}$, and $k=n / 2-\mathrm{t}$, we get:

$$
a_{n}=\frac{3 n^{2}+2 n-1}{2}, \quad \text { or } \quad a_{n}=\frac{3 n^{2}+4 n}{2}
$$

and using the principle from problem 954,

$$
a_{n}=\frac{3 n^{2}+\left[3+(-1)^{n}\right] n}{2}-\frac{1-(-1)^{n}}{4}=\frac{1}{4}\left[6 n^{2}+6 n-1+(2 n+1)(-1)^{n}\right]
$$

Margit Kovács (Szombathely, Savaria g. III. o. t.)

Remark: The formulas $a_{2 k-1}=6 k^{2}-4 k, a_{2 k}=6 k^{2}+4 k$ can also be derived from the following regularities, or from the requirement that they persist:

$$
\begin{array}{ccc}
a_{2}+a_{1}=12=12 \cdot 1^{2} & a_{4}+a_{3}=48=12 \cdot 2^{2} & a_{6}+a_{5}=108=12 \cdot 3^{2} \\
a_{2}-a_{1}=8=8 \cdot 1 & a_{4}-a_{3}=16=8 \cdot 2 & a_{6}-a_{5}=24=8 \cdot 3 \\
a_{8}+a_{7}=192=12 \cdot 4^{2} & a_{10}+a_{9}=300=12 \cdot 5^{2} \\
a_{8}-a_{7}=32=8 \cdot 4 & a_{10}-a_{9}=40=8 \cdot 5
\end{array}
$$

Thus, we can take the following as the rule for further development:

$a_{2 k}+a_{2 k-1}=12 k^{2}, a_{2 k}-a_{2 k-1}=8 k$, and from these, as a system of equations, $a_{2 k}$ and $a_{2 k-1}$ can be derived.

II. Solution: The difference sequence $d_{k}$ of the (odd) sequence has a similar structure to the sequence seen in problem 954:

$$
\begin{aligned}
& d_{2 k-1}=8+8(k-1)=8 k, \quad d_{n}=4 n+4, \quad \text { if } n \quad \text { is odd, } \\
& d_{2 k}=6+4(k-1)=4 k+2, \quad d_{n}=2 n+2, \quad \text { if } n \quad \text { is even. }
\end{aligned}
$$

For the purpose of expressing $a_{n+1}=a_{1}+\left(d_{1}+d_{2}+\ldots+d_{n}\right)$, let's also write the sum of the first $n$ terms of the $d$-sequence in the manner of problem 954

$$
S_{2 k}=S_{k}^{\prime}+S_{k}^{\prime \prime}=\frac{k}{2}\left(8+d_{2 k-1}+6+d_{2 k}\right)=\frac{k}{2}(16+12 k)=6 k^{2}+8 k
$$

$S_{2 k-1}=S_{2 k}-d_{2 k}=6 k^{2}+4 k-2$, that is,

$$
S_{n}=\left\{\begin{array}{lll}
\frac{3 n^{2}+8 n}{n^{2}+10 n+3}, & \text { if } n \text { is even, } \\
\frac{3 n^{2}+10}{2}, & \text { if } n \text { is odd. }
\end{array}\right.
$$[^0]

Summarizing: for any positive integer value of $n$.

$$
S_{n}=\frac{1}{4}\left[6 n^{2}+18 n+3-(2 n+3)(-1)^{n}\right],
$$

and thus:

$$
a_{n+1}=2+S_{n}=\frac{1}{4}\left[6 n^{2}+18 n+11-(2 n+3)(-1)^{n}\right]
$$

finally, substituting $n-1$ for $n$, and $-(-1)^{n}$ for $(-1)^{n-1}$,

$$
a_{n}=\frac{1}{4}\left[6(n-1)^{2}+18(n-1)+11-(2 n+1)(-1)^{n-1}\right]=\frac{1}{4}\left[6 n^{2}+6 n-1+(2 n+1)(-1)^{n}\right] .
$$

Nándor Bürger (Budapest, József A. g. III. o. t.)


[^0]:    ${ }^{1}$ See p. 106 of this issue.