Consider a board with 11 rows and 11 columns.

![](https://cdn.mathpix.com/cropped/2024_05_01_d889f7c81233a28648eeg-5.jpg?height=349&width=349&top_left_y=408&top_left_x=682)

Figure 29.1

(a) How many cells form this board?

(b) The diagonal whose cells are shaded separates the board into two regions: one above and one below. How many cells form each region? Is it possible to calculate this number without counting each cell?

(c) With the help of the board, it is possible to calculate the sum $1+2+\cdots+10$. Explain how.

(d) With the help of another board, using a similar reasoning to the previous item, it is possible to calculate the sum $1+2+\cdots+100$. What should be the number of rows and columns of the board? What is the value of the sum?
(a) Since there are 11 squares in each row of the board and it has 11 rows, the total number of squares is $11 \times 11=121$.

(b) Since there is one square of the diagonal in each row of the board and it has 11 rows, the total number of squares on the diagonal is 11. On the other hand, the diagonal is an axis of symmetry, dividing two equal regions. There are $11 \times 11$ squares on the board, and 11 of these are on the diagonal. The number of squares that form each region is then $(11 \times 11-11) / 2=55$.

(c) Let's count the number of squares in each piece by row (see Figure 29.2). The first row contains 1 square, the second 2, the third 3, and so on, up to the last row, which contains 10 squares. Therefore, the sum $1+2+\cdots+10$ is the total number of squares in each piece, which contains 55 squares:

$$
1+2+\cdots+10=\frac{11 \times 11-11}{2}=55
$$

Suggestion: Observe that the two regions formed are equal. In item (c), count the squares of each piece by row.

![](https://cdn.mathpix.com/cropped/2024_05_01_d889f7c81233a28648eeg-5.jpg?height=363&width=331&top_left_y=1977&top_left_x=1502)

(d) Let's consider a board with 101 rows and 101 columns and consider the diagonal that divides it into two equal regions. The diagonal contains 101 squares and each region contains $(101 \times 101-101) / 2=5050$ squares. On the other hand, counting the number of squares by row, we get $1+2+\cdots+100$. Therefore,

$$
1+2+\cdots+100=5050
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_d889f7c81233a28648eeg-6.jpg?height=343&width=388&top_left_y=314&top_left_x=206)

Figure 29.3

## Related Problem

Consider the board with 10 rows and 10 columns, as shown in Figure 29.3. It is divided into ten pieces in the shape $\square$ alternately colored white and gray. The first piece is formed by only one square.

(a) How many squares form the seventh piece? And the tenth piece?

(b) Is it possible to calculate the sum $1+3+\cdots+19$ with the help of this board? How?

(c) With a similar reasoning and the help of another board, it is possible to calculate the sum $1+3+5+\cdots+99$. How many rows and columns should the board have? What is the value of the sum?