Question

For Problems 45–50, use a graphing utility to find the sum of each geometric sequence.

$\frac{1}{4}+\frac{2}{4}+\frac{2^2}{4}+\frac{2^3}{4}+...+\frac{2^{14}}{4}$

Short answer

The sum of the given sequence is$8191.75$

Step-by-step solution

Step 1. Write the given information.

The given sequence is$\frac{1}{4}+\frac{2}{4}+\frac{2^2}{4}+\frac{2^3}{4}+...+\frac{2^{14}}{4}$

Step 2. Find the common ratio.

$$\begin{gathered} a_1=\frac{1}{4},a_2=\frac{2}{4},n=15 \\ \frac{a_2}{a_1}=\frac{{\displaystyle \frac{2}{4}}}{{\displaystyle \frac{1}{4}}} \\ =2 \\ r=2 \end{gathered}$$

Step 3. Find the sum.

The formula for the sum of nth term is $s_n=\frac{a_1(r^n-1)}{r-1}$

$$\begin{gathered} s_{15}=\frac{{\displaystyle \frac{1}{4}}(2^{15}-1)}{2-1} \\ =\frac{32768-1}{4} \\ =8191.75 \end{gathered}$$