Why use induction?
Theorem: The sum of the first $n$ powers of two is $2^n-1$.
Proof: Let $P(n)$ be the statement “the sum of the first $n$ powers of two is $2^n-1$.” We will prove, by induction, that $P(n)$ is true for all $n \in \mathbb N,$ from which the theorem holds.
For our base case, we need to show that $P(0)$ is true, meaning that the sum of the first 0 powers of two is $2^0-1.$ Since the sum of the first 0 powers of two is 0 and $2^0-1=0$ as well, we see that $P(0)$ is true.
For the inductive step, assume for some arbitrary $k \in \mathbb N$ that $P(k)$ holds, meaning that
$$ 2^0 + 2^1 + \cdots + 2^{k-1}=2^k-1. $$
We need to show that $P(k+1)$ holds, meaning that the sum of the first $k+1$ powers of two is $2^{k+1}-1.$ To see this, notice that
$$ \begin{align*} 2^0 + 2^1 + \cdots + 2^{k-1} + 2^k &= (2^0+ \cdots + 2^{k-1}) + 2^k\\ &= (2^k-1)+2^k\\ &= 2(2^k) - 1\\ &= 2^{k+1}-1. \end{align*} $$
Therefore, $P(k+1)$ is true, completing the induction. $\blacksquare$
Problem statement: 3 seemingly identical coins, 2 of which are real and one of which is counterfeit. The counterfeit coin is heavier. How to determine which is the counterfeit with one weighing of the balance?
9 identical coins, 1 heavier counterfeit, 2 weights