# Geometry Pattern Formula

• Sep 3rd 2008, 02:17 PM
Rocker1414
Geometry Pattern Formula
Hello All, thank you for your help.

I am at a complete loss in this problem:

Attatched are the triangular numbers used to do the problem, and a table clarifying a few things.

The "n" row on the table states what n equals.
The "nth Triangular Number" row states how many dots there are when n=a certain number. (So when n=1, there is one dot.)

I am to find a formula for the nth triangular number.

The book says that each triangular number could be thought of as half the area of a rectangle whose width is the same as the triangle number, and it's height is n+1. So if n=2, the box is thought of as 2 units wide and 3 units tall, for a total of 6 units.

It also says that the formula for this problem is:
S-1/2(Width)(Length)=1/2(n)(n+1)

I have no idea how one would arrive at such an explanation for a pattern of said triangular numbers, nor how one would arrive at such a formula.

Any help on the correct way to solve a formula such as this and how to find the formula would be most appreciated.
Thanks a lot,
Rocker1414
• Sep 3rd 2008, 03:12 PM
skeeter
sequence of triangular numbers ...

1, 3, 6, 10, 15, ...

double them ...

2, 6, 12, 20, 30, ...

factor each term ...

1*2, 2*3, 3*4, 4*5, 5*6, ... , n(n+1)

undouble (halve) ...

1, 3, 6, 10, 15, ... , n(n+1)/2
• Sep 3rd 2008, 07:34 PM
Soroban
Hello, Rocker1414!

Quote:

Attatched are the triangular numbers used to do the problem, and a table clarifying a few things.

The $n$ row on the table states what n equals.
The $T_n$ row states how many dots there are when $n$ = a certain number.

I am to find a formula for the $n^{\text{th}}$ triangular number.

$\begin{array}{c|cccc} n &\quad 1 & \quad2 & \quad3 & \quad4 \\ \hline
& & & & \quad\bullet \\

\end{array}$

That hint goes like . . .

Suppose we want to know $T_5$ without counting them.

We have: . $\begin{array}{c} \bullet \\ \bullet\;\bullet \\ \bullet\;\bullet\;\bullet \\ \bullet\;\bullet\;\bullet\;\bullet \\ \bullet\;\bullet\;\bullet\;\bullet\;\bullet\end{ar ray}$

Left-justify the dots: . $\begin{array}{cccccc} \bullet & & & & \\
\bullet & \bullet & & & \\
\bullet & \bullet & \bullet & & \\
\bullet & \bullet & \bullet & \bullet & \\
\bullet & \bullet & \bullet & \bullet & \bullet\end{array}$

Add a mirror-image copy of the triangle: . $\begin{array}{cccccc}\bullet& \circ&\circ&\circ&\circ&\circ \\
\bullet&\bullet&\circ&\circ&\circ&\circ \\
\bullet &\bullet& \bullet & \circ & \circ & \circ \\
\bullet & \bullet&\bullet&\bullet&\circ&\circ \\
\bullet & \bullet & \bullet & \bullet & \bullet & \circ\end{array}$

We have a 5-by-6 rectangle with a total of: . $5 \times 6 \:=\:30$ dots.

The triangle contains half that many dots.

Therefore: . $T_5 \;=\;\frac{5\cdot6}{2} \;=\; 15$

In general: . $\boxed{T_n \;=\;\frac{n(n+1)}{2}}$