# Exercise 2.71

Tree for \(n = 5\) with only the weight on each node and leaf: {: .center-image .medium-margin} Tree for \(n = 10\) with only the weight on each node and leaf: {: .center-image .medium-margin} In such a tree only one bit is required to encode the most frequent symbol and \(n-1\) bits are required for the least frequent symbol. The shape of the tree come from the fact that \(2^{n-1}>\sum\limits_{i=0}^{n-2} 2^i\). To prove this we can see that \(\sum\limits_{i=0}^{n-2} 2^i=2^{n-1}-1\) (we can easily prove this by induction).