The first three patterns in a sequence of patterns are shown below - Leaving Cert Mathematics - Question 7 - 2015

To show that the numbers of black triangles are quadratic, we first list the known counts: 3, 6, 10, 15, 21.

We find the first difference:

• 6 - 3 = 3
• 10 - 6 = 4
• 15 - 10 = 5
• 21 - 15 = 6

Now the second differences:

• 4 - 3 = 1
• 5 - 4 = 1
• 6 - 5 = 1

Since the second difference is constant, the sequence is quadratic.

The black triangles follow the formula

$B_n = \frac{1}{2}n^2 + 3n$

For n = 9:

$B_9 = \frac{1}{2}(9^2) + 3(9) = \frac{1}{2}(81) + 27 = 40.5 + 27 = 55$

Thus, there are 55 black triangles in the 9th pattern.

Using the earlier calculation, the total number of triangles in the 9th pattern is:

$T_n = (n + 1)^2$

For n = 9:

$T_9 = (9 + 1)^2 = 10^2 = 100$

The number of white triangles:

$W_n = T_n - B_n = 100 - 55 = 45$

So, there are 45 white triangles in the 9th pattern.

Using the formula for the total number of triangles:

$T_n = (n+1)^2$ for n = 9, gives us $T_9 = 100$.

Thus, in the 9th pattern, there are 100 small triangles in total.

The expression for the total number of triangles in the nth pattern is given by:

$T_n = (n + 1)^2$

The formula for black triangles is:

$B_n = \frac{1}{2}n^2 + 3n + c$

If we plug in n = 1:

$B_1 = 1 \Rightarrow \frac{1}{2}(1^2) + 3(1) + c = 1\n\Rightarrow\frac{1}{2} + 3 + c = 1 \Rightarrow\ c = 1 - 3.5 = -2.5$

Given:

$W_n + B_n = T_n$

Thus,

$W_n = T_n - B_n \Rightarrow W_n = (n + 1)^2 - \left(\frac{1}{2}n^2 + 3n - 2.5\right)$

This simplifies to:

$W_n = \frac{1}{2}n^2 + n + 2.5$

For the pattern having a total of 625 triangles:

$T_n = 625 = (n + 1)^2 \Rightarrow n + 1 = 25 \Rightarrow n = 24$

Now, for black triangles:

$B_{24} = \frac{1}{2}(24)^2 + 3(24) + c = 288 + 72 - 2.5 = 357.5$

Number of white triangles:

$W_{24} = 625 - 357.5 = 300$

Thus, there are 325 black triangles and 300 white triangles.