Formula for the triangular numbers


Simplify \(T(n - 1)\)


$$\begin{align} T(n - 1) &= \dfrac{(n - 1)[(n - 1) + 1]}{2} \\[0.5em] &= \dfrac{n(n - 1)}{2} \end{align}$$

Note that this exercise is useful, because it's part of the derivation of one of the formulas for the polygonal numbers.


In the multiplication square below, the first eleven triangle numbers have been highlighted. Use this pattern to predict the next few triangular numbers, then check to see if you're right. Why does this pattern hold?



Are there more two-digit numbers with the first digit larger than the second, than there are with the second digit larger than the first?

Solution 1:

If the first digit is 1, then the two-digit numbers where the first digit is greater than the second includes 10 only. If the first digit is 2, then we have 20 and 21. If 3, then 30, 31, and 31. Thus, we can recognize the triangular numbers. More specifically, there will be

$$T_9 = \dfrac{9(9 + 1)}{2} = \dfrac{9 \cdot 10}{2} = 9 \cdot 5 = 45$$

numbers such that the first digit is greater than the second. This is half the number of two-digit numbers. But we know from the remaining half, some numbers will have their first digit equal to their second, namely, 11, 22, 33, and so on. There are 9 such numbers. Thus, the number of two-digit numbers where the second digit is greater than the first, must be 45 - 9 = 36.

Solution 2:

This solution is much more clever. Have a look at a hundreds chart, or make a table of the two-digit numbers, as seen below. Draw a diagonal line through the numbers where the first digit is equal to the second. You'll see right away there are more numbers below the diagonal than above, and all the numbers below the diagonal are the numbers where the first digit is greater than the second.

Going further:

What about in other bases?


This problem comes from "The Inquisitive Problem Solver," problem 23, part a. I added another solution and extended the problem.