Consider the following sequence of pieces, where the piece numbered 1 is a small square.
![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-12.jpg?height=162&width=380&top_left_y=1508&top_left_x=677)

(a) How many small squares form the piece numbered 50?

(b) How many small squares are there in the union of the pieces numbered 1 to 50?

(c) Observing the result from part b, calculate

$$
2+4+6+\ldots+96+98+100
$$

(d) Calculate

$$
1+2+3+4+\ldots+97+98+99+100
$$
Solution

(a) See that with each increment of one unit in the piece number, the number of squares increases by 2 units. Therefore, piece 50 will have $1+2 \cdot 49=99$ squares.
(b) See that the pieces can be juxtaposed to form a larger square divided into squares. What determines the side of this square is the number of the largest piece used. Thus, the 50 pieces have a total of $50 \cdot 50=2500$ squares.

(c) See that the previous item provides us with the equation:

$$
1+3+5+\ldots+95+97+99=2500
$$

If we add 1 to each of the 50 terms of the previous sum, we will get the desired sum:

$$
\begin{aligned}
1+3+5+\ldots+95+97+99 & =2500 \\
(1+1)+(3+1)+(5+1)+\ldots+(95+1)+(97+1)+(99+1) & =2500+50 \\
2+4+6+\ldots+96+98+100 & =2550
\end{aligned}
$$

(d) Taking into account the results obtained in the previous items, it is enough to add the sum of the odd numbers and the sum of the even numbers. Thus, we conclude that

$$
1+2+3+4+\ldots+97+98+99+100=2500+2550=5050
$$