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. the formula for the sum of a finite geometric series of n terms with first term $t_1$ and common ratio r is $s_n = t_1left(\frac{1 - r^n}{1 - r}
ight)$. after two tries, none of your answers are correct. try again.

Explanation:

Step1: Identify the first - term and common ratio

Let the area of the center triangle in the first figure be $t_1$. Assume the area of the large triangle in the first figure is $A$. If we consider the construction of the pattern, when we go from one figure to the next, the area of the new - formed center triangle is $\frac{1}{4}$ of the area of the center triangle of the previous figure. So, $t_1$ is the area of the center triangle in the first figure and the common ratio $r=\frac{1}{4}$.

Step2: Determine the number of terms

We want to find the sum of the areas of the center triangles in the first four terms of the sequence, so $n = 4$.

Step3: Use the geometric - series formula

The sum formula of a finite geometric series is $S_n=t_1\frac{1 - r^n}{1 - r}$. Substituting $n = 4$ and $r=\frac{1}{4}$ into the formula, we get $S_4=t_1\frac{1-(\frac{1}{4})^4}{1-\frac{1}{4}}$.
\[

$$\begin{align*} S_4&=t_1\frac{1-\frac{1}{256}}{\frac{3}{4}}\\ &=t_1\frac{\frac{256 - 1}{256}}{\frac{3}{4}}\\ &=t_1\frac{\frac{255}{256}}{\frac{3}{4}}\\ &=t_1\times\frac{255}{256}\times\frac{4}{3}\\ &=t_1\times\frac{85}{64} \end{align*}$$

\]
If we assume the area of the center triangle in the first figure $t_1 = 1$ (since the problem does not give the actual value of the first - term area, and we are interested in the relative sum), then $S_4=\frac{85}{64}$ square units.

Answer:

$\frac{85}{64}$