Going back once more to the original problem of finding a formula to compute the n triangular number.

$$T_n=1+2+3+ \cdots +n$$

Triangular numbers suggest by their name the idea of geometry and it should be interesting to try apply some geometric reasoning to solve the problem. As we explained in the first lecture triangular numbers are obtained when we arrange a number of stones in an equilateral triangular shape. It just happens that it is very convenient for us to re arrange those stones in another triangular shape. That is right angle triangle. In that way we are able to produce triangular numbers and it become very simple to arrange the stones in a two dimensional array of stones that could be easily counted. Now we will see how the geometric demonstration works for $T_4$.

First, we duplicate the number of stones so we have $2 T_4$ and conveniently rotating the second $T_4$ we can joint the first $T_4$ and we have as a result a rectangular shape.

The number of stones in that rectangular shape is very easy to compute. It will be the product of the number of stones in two sides. In this case $2 T_4=4(4+1)$ and since we have duplicated the original number $T_4 $ now we need to divide by $2$ and that will us the value $T_4=4(4+1)/2$. The same procedure could be carry out for $100$ stones or any other number. We do not suggest you do this literally for $100$ stones, but you should be able to play this same argument in your mind.

In the previous example we were extremely close to form a square number. We actually missed this by just one row. That is why we have the $(4+1)$ factor. A natural question to ask is.

What should we add to any triangular number $T_k$ to get a square number?

We can see it in the diagram. What we need is to add the prior triangular number! We can express this algebraically with the following relation

$$T_k+T_{k-1}=k^2$$

What this formula is saying is that the sum of two consecutive triangular numbers is an square number. We can verify that this is true since $T_1+T_{2}=1+3=4$ and also $T_2+T_{3}=3+6=9$. It is easy to see from the geometric configuration why the sum of two consecutive triangular numbers is an square. Since $T_k$ is something we are interested in finding then the relation seems to be also a kind of equation where the unknown to be found is $T_k$ . Notice also that $T_{k-1}$ appears on the equation. Naturally if we know how to compute $T_k$ we then also know to compute $T_{k-1}$. Equations of this type are known as recursive equations. The last row of $T_k$ will be the diagonal since subtracting the diagonal the resulting triangular number is equal to $T_{k-1}$. We can also write another relation

$$T_k=T_{k-1}+k$$

Now these two equations are defining $T_k$ in terms of the prior triangular number $T_{k-1}$. These type of relations are called recurrence equations and we will discussed them in more detail in some other lecture.

For now let us try and see if we can solve the first equation.

$$T_k+T_{k-1}=k^2$$

(This is a partial transcription of the Video Lecture)

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