Question

In Problems 39– 44, find each sum.

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

Short answer

The sum of the given sequence is$\frac{1}{4}·-(1-2^n)$

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.

$$\begin{gathered} a_1=\frac{1}{4},a_2=\frac{3}{4},a_3=\frac{2^2}{4} \\ \frac{a_2}{a_1}=\frac{{\displaystyle \frac{2}{4}}}{{\displaystyle \frac{1}{4}}} \\ =2 \\ \frac{a_3}{a_2}=\frac{{\displaystyle \frac{2^2}{4}}}{{\displaystyle \frac{2}{4}}} \\ =2 \\ r=2 \end{gathered}$$

Step 3. Find the sum of nth term.

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

$$\begin{gathered} s_n=\frac{1}{4}·\frac{1-2^n}{1-2} \\ =\frac{1}{4}·-(1-2^n) \end{gathered}$$