The diagram below shows the first 4 steps of an infinite pattern which creates the Sierpiński Triangle - Leaving Cert Mathematics - Question 9 - 2018

The number of black triangles in step n can be expressed as:

$3^n$

To find k where $3^k > 10^9$, we can take logarithms:

$k \log_{10}(3) > \log_{10}(10^9)$

This simplifies to:

$k > \frac{9}{\log_{10}(3)}$

Calculating:

$k > \frac{9}{0.477} \approx 18.863$

Thus, the smallest integer k is:

$k = 19$

To find h where ( \left( \frac{3}{4} \right)^h < \frac{1}{100} ), we can proceed with the following calculations:

$h \log_{10}\left(\frac{3}{4}\right) < \log_{10}\left(\frac{1}{100}\right)$

Substituting values:

$h > \frac{2}{\log_{10}\left(\frac{3}{4}\right)}$

Calculating:

This leads to:

$h \approx 17$

As n approaches infinity, the fraction of the original triangle remaining tends to:

$\lim_{n \to \infty} \left( \frac{3}{4} \right)^n = 0$

Thus, the fraction remaining is 0.

Using the formula for the perimeter, in step n the perimeter is given by:

$P_n = 3 + 6 \cdot 3^n$

For step 35, we calculate:

$P_{35} = 3 + 6 \cdot 3^{35}$

Exact value should be calculated and rounded to the nearest unit.

As n tends to infinity,

• The total area of the black triangles approaches 0 because the fraction remaining in the original triangle converges to 0.
• However, the total perimeter diverges to infinity since it continues to grow with the number of steps, indicating that perimeter increases while area decreases.