Let's find the regularities shown in the following table:

$$
\begin{aligned}
1 & =\quad 1 \\
2+3+4 & =1+8 \\
5+6+7+8+9 & =8+27 \\
10+11+12+13+14+15+16 & =27+64
\end{aligned}
$$

Write the common value of the two sides in the form of a product of two factors. Try to extend the table with additional rows. What will be the next row? Without writing the intermediate rows, give the first, middle, and last term on the left side of the twentieth row, as well as the two terms on the right side. Does the equality still hold there?
The written equalities are correct, with the two sides having a common value of 1, 9, 35, 91, respectively.

On the left side, there is the sum of consecutive integers, the last term of which is the square of consecutive numbers; however, the first term of each row is the next number after the last term of the previous row. Based on this, the next sum on the left side of the next row is: $17+18+19+\ldots+25$, the value of which is 189. On the right side, the sum of the cubes of two consecutive numbers stands, the larger of which is the cube of the number whose square ends the left side. Therefore, the sum $4^{3}+5^{3}=64+125=189$ should be written in the next row, which indeed results in an equality:

$$
17+18+19+20+21+22+23+24+25=64+125
$$

The common value of the two sides in the 2nd to 4th rows can be written as a product of two factors in only one way (1 is not allowed as a factor):

$$
3 \cdot 3, \quad 5 \cdot 7, \quad 7 \cdot 13
$$

In these factorizations, the first factor is the number of terms on the left side of the corresponding row, and the second factor is the middle term in terms of magnitude. The common value of the two sides in the 5th row can be written as a product of two factors in three ways:

$$
189=3 \cdot 63=7 \cdot 27=9 \cdot 21
$$

Among these, the last one repeats the observed regularity.

Based on the regularity followed in forming the 5th row, the left side of the 20th row will contain the sum of integers from $19^{2}+1=362$ to $20^{2}=400$. The number of terms is $400-361=39$, and the 20th term, which is the middle one, is 381. The sum on the right side will be $19^{3}+20^{3}$.

The sum on the left side can be calculated based on the fact that the sum of a term preceding 381 and the term following it is always $2 \cdot 381$, because the preceding term is as much smaller than 381 as the following term is larger. 19 such pairs can be formed; adding the middle term, we get a total of $2 \cdot 19+1=39$ times the middle term, 381. Thus, the value of the sum is $39 \cdot 381=14859$. From the sum of the two cubes on the right side, by factoring out the sum of the bases, we get that $19^{3}+20^{3}=(19+20) \cdot\left(19^{2}-19 \cdot 20+20^{2}\right)=39 \cdot\left[19^{2}+20(-19+20)\right]=39 \cdot 381=14859$.

Thus, the equality also holds in the 20th row.

The thought process followed here is also suitable for demonstrating that the equality holds for any row written according to the found regularity, and the common value of the two sides can be written as the product of the number of terms on the left side and the middle term.

Baranyai Zsolt (Budapest, Rákóczi F. Gymnasium II. o. t.)