Find height of a ternary tree Ternary heap is like a binary tree, just every node can have up to $3$ sons and not $2$.

I try to bound the number of nodes in the heap, $n$, using the height of the heap $h$.

The solutions get to: 

$$
3^h &lt; n &lt; 3^{h+1}
$$

Yet, I get to: 

$$
\frac{3^h}{2} &lt; n &lt; \frac{3^{h+1}}{2}
$$

In short, what I do is: 

$$
\sum_{i = 0}^{h - 1}3^i = \frac{3^h-1}{2}
$$

For all the nodes in all levels except for the last level, because all the levels are full. 

If the last level is full: 

$$
\frac{3^h - 1}{2} + 3^h
$$

If the last level has only $1$ node, we get: 

$$
\frac{3^h - 1}{2} + 1
$$

From here I conclude what I showed at the beginning. 

Why the solutions get to something else?
The expected solution is wrong.

Pick $h=0$. The only ternary heap with height $0$ has only one node. Nevertheless the expected solution says that it needs to have at least $3^0 + 1 = 2$ nodes.

In case you meant to write $3^h \le n \le 3^{h+1}$ instead of $3^h &lt; n &lt; 3^{h+1}$, the expected solution is still wrong.
Consider a ternary heap with $n=2$ nodes. This heap has height $1$ but according the expected solution it should have at least $3^h = 3$ nodes.