Is this a good way of demonstrating the formula for geometric series?

I don’t know if the formula for geometric series is taught in high school anymore, but it is a rather important equation in mathematics, and it can be derived without algebraic manipulation if we limit ourselves to the cases for positive whole numbers. We want to find an equation for 1 + a + a^{2 + a}3 + ... +a^n. The explanation is a bit lengthy but should be easy to follow.

Suppose we have a tennis tournament for 2^5 = 32 players. The first round of the tournament will have 16 games. The succeeding rounds will have 8, 4, 2 and 1 games. What is an easy way of finding the total number of games?

Each game eliminates one player. In total, 31 players must be eliminated to have one tournament winner. There must therefore be 31 games to eliminate the 31 players, so 16 + 8 + 4 + 2 +1 = 31.

Let’s see if we can generalize this. Suppose we have some game that allows for 3 players, like the board game of Monopoly. If we start with 3^5 = 243 players, the first round will have 243/3 = 81 games and the total number of games is 81 + 27 + 9 + 3 + 1.

This time 2 players are eliminated in each game and we have to eliminate 80 players. There must be 80/2 = 40 games.

If a is any positive whole number, to find the value of 1 + a + a^{2 + ... + a}n, Imagine a tournament with a^{(n+1) players. The preceding equation gives the total number of games. (a – 1) players are eliminated per game and there are a}(n+1) – 1 players that must be eliminated, so the total number of games is
[a^{(n+1)-1] /(a-1). We have 1 + a + a}2 + ... + a^{n = [a}(n+1)-1] /(a-1).