Question

find the first four partial sums of the series.\\(\frac{1}{4} - \frac{1}{16} + \frac{1}{64} - \frac{1}{256} + \frac{1}{1024} - \frac{1}{4096} + \dots\\) write your answers as integers or fractions in simplest form.\\(s_1 = \square\\)\\(s_2 = \square\\)\\(s_3 = \square\\)\\(s_4 = \square\\)

Explanation:

Step1: Find \( S_1 \)

The first partial sum \( S_1 \) is just the first term of the series. The first term is \( \frac{1}{4} \), so \( S_1=\frac{1}{4} \).

Step2: Find \( S_2 \)

The second partial sum \( S_2 \) is the sum of the first two terms. The first term is \( \frac{1}{4} \) and the second term is \( -\frac{1}{16} \). So we calculate \( \frac{1}{4}+(-\frac{1}{16})=\frac{4}{16}-\frac{1}{16}=\frac{3}{16} \), thus \( S_2 = \frac{3}{16} \).

Step3: Find \( S_3 \)

The third partial sum \( S_3 \) is the sum of the first three terms. We already know \( S_2=\frac{3}{16} \), and the third term is \( \frac{1}{64} \). So \( S_3=S_2 + \frac{1}{64}=\frac{3}{16}+\frac{1}{64}=\frac{12}{64}+\frac{1}{64}=\frac{13}{64} \).

Step4: Find \( S_4 \)

The fourth partial sum \( S_4 \) is the sum of the first four terms. We know \( S_3 = \frac{13}{64} \), and the fourth term is \( -\frac{1}{256} \). So \( S_4=S_3+(-\frac{1}{256})=\frac{13}{64}-\frac{1}{256}=\frac{52}{256}-\frac{1}{256}=\frac{51}{256} \).

Answer:

\( S_1 = \frac{1}{4} \)
\( S_2 = \frac{3}{16} \)
\( S_3 = \frac{13}{64} \)
\( S_4 = \frac{51}{256} \)