Answer
\[
\boxed{
t_n =
\begin{cases}
1, & n=0,\\
1, & n=1,\\
2, & n=2,\\
3, & n=3,\\
t_{n-1} + t_{n-2} + t_{n-4}, & n\ge 4.
\end{cases}
}
\]
That is the desired recurrence relation for \(t_1, t_2, t_3, \dots\).
*(If one prefers to specify only for \(n\ge 1\), you would keep in mind that \(t_{n-4}=0\) when \(n-4<0\). But the piecewise definition above is often simplest.)*
Work Step by Step
Below is a common way to derive a recurrence for the number of ways to build a tower of height \(n\) using blocks of heights 1 cm, 2 cm, and 4 cm, assuming an unlimited supply of each.
---
## 1. Define \(t_n\)
Let \(t_n\) be the number of ways to construct a tower of total height \(n\) (in centimeters) using blocks of heights 1, 2, or 4 cm. By “ways,” we mean different sequences of blocks stacked on top of each other that sum to height \(n\).
### Initial Conditions
- **\(t_0\)**: There is exactly 1 way to make a “tower” of height 0—by using no blocks at all. So \(t_0 = 1.\)
- **\(t_1\)**: To get height 1, you must use a single 1 cm block, so \(t_1 = 1.\)
- **\(t_2\)**: To get height 2, either use two 1 cm blocks (1+1) or one 2 cm block, so \(t_2 = 2.\)
- **\(t_3\)**: For height 3, you can build it from height 2 by adding a 1 cm block, or from height 1 by adding a 2 cm block, but you cannot add a 4 cm block (that would overshoot 3). Enumerating quickly:
1. 1 cm + 1 cm + 1 cm
2. 1 cm + 2 cm
3. 2 cm + 1 cm
Hence \(t_3 = 3.\)
*(For \(n<0\), we take \(t_n=0\) by convention, since you can’t build a negative‐height tower.)*
---
## 2. Recurrence Relation
To build a tower of height \(n \ge 4\):
1. **Top block is 1 cm:** Then the rest of the tower must be a valid tower of height \(n-1\). There are \(t_{n-1}\) ways to build that.
2. **Top block is 2 cm:** Then the rest of the tower must be a valid tower of height \(n-2\). There are \(t_{n-2}\) ways for that.
3. **Top block is 4 cm:** Then the rest of the tower must be a valid tower of height \(n-4\). There are \(t_{n-4}\) ways for that.
These three cases are disjoint and cover all possibilities. So the total number of ways to get height \(n\) is
\[
t_n \;=\; t_{n-1} \;+\; t_{n-2} \;+\; t_{n-4},
\quad\text{for }n \ge 4.
\]
