Question

suppose this pattern continues. the sum of the areas of the center triangles in the first four terms of this sequence of figures is square units.

Explanation:

Step1: Assume initial triangle area

Let the area of the largest triangle in the first - figure be \(A = 1\) square unit. In the second figure, the center - triangle has an area that is \(\frac{1}{4}\) of the area of the triangle in the first figure. In the third figure, the center - triangle has an area that is \((\frac{1}{4})^2\) of the area of the triangle in the first figure. In the fourth figure, the center - triangle has an area that is \((\frac{1}{4})^3\) of the area of the triangle in the first figure.

Step2: Calculate the sum of areas

The sum \(S\) of a geometric series with first term \(a\), common ratio \(r\), and \(n\) terms is given by \(S=\sum_{k = 0}^{n - 1}ar^k=\frac{a(1 - r^n)}{1 - r}\) (when \(r
eq1\)). Here, \(a = 1\), \(r=\frac{1}{4}\), and \(n = 4\). So \(S = 1\times\frac{1-( \frac{1}{4})^4}{1-\frac{1}{4}}\).
\[

$$\begin{align*} S&=\frac{1-\frac{1}{256}}{\frac{3}{4}}\\ &=\frac{\frac{256 - 1}{256}}{\frac{3}{4}}\\ &=\frac{\frac{255}{256}}{\frac{3}{4}}\\ &=\frac{255}{256}\times\frac{4}{3}\\ &=\frac{255}{192}\\ &=\frac{85}{64} \end{align*}$$

\]

Answer:

\(\frac{85}{64}\)