Which is the arithmetic sequence in which the sum of any number of terms is always twice the square of the number of terms?
I. solution. We can give the sequence without knowing the sum formula. Indeed, for $n=1$ and 2,

\[
\begin{aligned}
& S_{1}=a_{1}=4 \cdot 1^{2}=4 \text{, so } a_{1}=4 \\
& S_{2}=a_{1}+a_{2}=4+a_{2}=4 \cdot 2^{2}=16 \text{, so } a_{2}=12
\end{aligned}
\]

thus $d=a_{2}-a_{1}=8$, and the sequence is: $4,12,20,28,36, \ldots$

The requirement is satisfied for all $n$, because in this case $a_{n}=4+8(n-1)=8 n-4$, and the sum

\[
S_{n}=\frac{1}{2}\left(a_{1}+a_{n}\right) n=\frac{1}{2}(4+8 n-4) n=4 n^{2}
\]

Endreffy Zoltán (Budapest, I. István g. III. o. t.)

# II. solution. 

\[
a_{n}=S_{n}-S_{n-1}=4 n^{2}-4(n-1)^{2}=8 n-4
\]

And from this, for $n=1$ and 2, $a_{1}=4, a_{2}=12$, so $d=8$.

Kovács Imre (Békés, Szegedi Kis I. g. IV. o. t.)

III. solution. Based on the sum formula, the requirement

\[
S_{n}=\frac{1}{2}\left[2 a_{1}+(n-1) d\right] n=4 n^{2}
\]

from (simplifying with non-zero $n$) after rearrangement gives:

\[
(d-8) n+\left(2 a_{1}-d\right)=0
\]

This can only hold for all (positive integer) $n$ if the left side is identically 0:

\[
d-8=0, \quad 2 a_{1}-d=0
\]

thus $d=8$, and $a_{1}=d / 2=4$.