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# Practice Paper 4 Question 11

You have a binary tree with $$n$$ levels similar to the figure shown. On each level you draw a line passing through all nodes on that level. What is the total number of triangles formed in this way? Give a recursive expression for this number.

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## Warm-up Questions

1. Differentiate $$y=x^n.$$
2. Evaluate the sum $$\sum_{k=0}^{n}{2^k}.$$ Note: This result is important in Computer Science.
3. A geometric progression starts with $$x, 2x^2, 4x^3, \ldots.$$ Find the recursive formula of the sequence.

## Hints

• Hint 1
How many possible triangles have their top tip on the root node?
• Hint 2
Could you then find an expression for the number of possible triangles with the top tip on a node at level $$k.$$
• Hint 3
How many nodes are there on level $$k?$$
• Hint 4
Try to express the sum of the triangles at each level into two sums, one of which is easier to calculate.
• Hint 5
To calculate the other sum, find a general expression in terms of $$x$$ whose derivative has a similar form to our sum.
• Hint 6
You should find the general expression easier to simplify. Take the derivative of the simplified expression.
• Hint 7
Which value should $$x$$ take so that we get the result of our sum?
• Hint 8
To find a recursive expression, think of how you could build the $$(n+1)$$-level binary tree from the $$n$$-level binary tree.
• Hint 9
Have you tried building the 3-level binary tree from the 2-level binary tree?

## Solution

Every node at level $$k$$ has $$n-k$$ possible triangles with the top tip on that node. There are $$2^{k-1}$$ nodes at level $$k$$. Hence $$t_n=\sum_{k=1}^{n-1} (n-k)2^{k-1}$$$$=n(2^{n-1}-1)-\sum_{k=1}^{n-1}k2^{k-1}.$$ This last sum can be solved by differentiating for $$\sum_k x^k$$ and substitute $$x=2.$$ We get $$t_n=n2^{n-1}-n-n2^{n-1}+2^n-1$$$$=2^n-n-1.$$

To work out a recursive expression for this, realise that we can build the $$(n+1)$$-level binary tree (with horizontal lines) from the $$n$$-level one by making another copy horizontally, and then adding a root node at the top. Copying the existing tree would double the number of existing triangles, and then adding a root node at the top adds $$n$$ new triangles that didn't exist before. Hence $$t_{n+1} = 2t_n + n.$$

Alternatively we can subtract consecutive terms to get $$t_{n+1} = t_n + 2^n - 1.$$

Note: It's also possible to figure out the answer by progressive trials and then prove it by induction.

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