Answer
For \(n \ge 1\):
- \(c_1 = 1\),
- \(c_2 = 2\),
- and for \(n \ge 3\), \(c_n = c_{n-1} + c_{n-2}\).
This is essentially the same recursion as the Fibonacci sequence, shifted slightly in indexing.
Work Step by Step
## 1. Define \(c_n\)
For each integer \(n \ge 1\), let \(c_n\) be the number of different ways to climb a staircase of \(n\) stairs if, on each move, you can go up either 1 stair or 2 stairs.
### Initial Conditions
- **\(n=1\)**: There is exactly 1 way to climb 1 stair (just take a single 1‐stair step), so \(c_1 = 1.\)
- **\(n=2\)**: There are 2 ways to climb 2 stairs: either (1,1) or (2). Hence \(c_2 = 2.\)
---
## 2. Recurrence Relation
Consider a staircase of \(n\) stairs (with \(n \ge 3\)):
1. **Look at the first move you take**:
- If your first move is **1 stair**, then you still have \((n-1)\) stairs left to climb. By definition, there are \(c_{n-1}\) ways to climb those remaining \((n-1)\) stairs.
- If your first move is **2 stairs**, then you have \((n-2)\) stairs left. There are \(c_{n-2}\) ways to climb those remaining \((n-2)\) stairs.
2. **Sum of possibilities**:
These two cases (first step of size 1 vs. first step of size 2) are disjoint and cover all possibilities. Therefore, the total number of ways to climb \(n\) stairs is
\[
c_n = c_{n-1} + c_{n-2}.
\]
Hence the sequence \(\{c_n\}\) satisfies the linear recurrence
\[
\boxed{c_n = c_{n-1} + c_{n-2} \quad\text{for }n \ge 3,}
\]
with the initial conditions
\[
\boxed{c_1 = 1,\quad c_2 = 2.}
\]
