In the pattern below, we add a new row to the bottom of the triangle with every step. The next term will then be $10+5=15,$ and we can find the one after that by adding six: $15+6=21$.

We could extend this pattern as far as we want, making larger and larger triangles, but is it possible to skip straight to the number of dots in the $100^\text{th}$ triangle without calculating all the figures in between?

To find the number of dots in the $100^\text{th}$ triangle, we should look for an explicit formula that describes the number of dots in each figure. However, it’s not immediately obvious how we might describe the number of dots $($apart from the already established pattern of $1 + 2 + 3 + 4 + \cdots).$

If we rearrange and *double* the triangles, though, it’s possible to make each of them into a rectangular array, which we can then describe algebraically.

After we’ve doubled the triangles, each triangle forms a rectangle with these dimensions: one side has the *same* number of dots as the base of the triangle, while the other side is *one unit longer* than that.

Since the base of each triangle is $n$ dots, these rectangles have $n(n + 1)$ dots, so the triangles (which are half) have $\frac{n(n + 1)}{2}$ dots.

By rearranging dots, we've made it easier to find a general formula for the number of dots in the $n^\text{th}$ term, which is $\frac{n(n + 1)}{2}$ dots. We can determine that the $100^\text{th}$ term in the sequence has $\begin{aligned} \frac{100(100 + 1)}{2} & = \frac{100\times 101}{2} \\ & = \frac{10,100}{2} \\ & = 5,050 \\ \end{aligned}$ dots.

Can you find the pattern in the dots below?