Question

1. find the first four partial sums of the series.

s = 16 + 32 + 64 + 128 + 256 + ...

1. write your answers as integers or fractions in simplest form.

s₁ =
s₂ =
s₃ =
s₄ =

Explanation:

Step 1: Find \( S_1 \)

The first partial sum \( S_1 \) is just the first term of the series. The series starts with 8 (wait, the series is written as \( 8 + 16 + 32 + 64 + 128 + 256 + \dots \)? Wait, the initial expression: " \( 8 = 16 + 32 + 64 + 128 + 256 + \dots \)" – maybe a typo, should be \( 8 + 16 + 32 + 64 + 128 + 256 + \dots \). So first term \( a_1 = 8 \), so \( S_1 = 8 \).

Step 2: Find \( S_2 \)

The second partial sum \( S_2 \) is the sum of the first two terms: \( a_1 + a_2 = 8 + 16 = 24 \).

Step 3: Find \( S_3 \)

The third partial sum \( S_3 \) is the sum of the first three terms: \( a_1 + a_2 + a_3 = 8 + 16 + 32 = 56 \).

Step 4: Find \( S_4 \)

The fourth partial sum \( S_4 \) is the sum of the first four terms: \( a_1 + a_2 + a_3 + a_4 = 8 + 16 + 32 + 64 = 120 \).

Wait, wait, maybe the series is \( 8 + 16 + 32 + 64 + 128 + 256 + \dots \)? Let's check again. The problem says "Find the first four partial sums of the series. \( 8 + 16 + 32 + 64 + 128 + 256 + \dots \)" (assuming the "=" was a typo for "+"). So:

\( S_1 = 8 \)

\( S_2 = 8 + 16 = 24 \)

\( S_3 = 8 + 16 + 32 = 56 \)

\( S_4 = 8 + 16 + 32 + 64 = 120 \)

Wait, but let's confirm the series. If the first term is 8, common ratio 2 (since 16/8=2, 32/16=2, etc.). So partial sums:

\( S_1 = 8 \)

\( S_2 = 8 + 16 = 24 \)

\( S_3 = 24 + 32 = 56 \)

\( S_4 = 56 + 64 = 120 \)

Yes, that makes sense.

Answer:

\( S_1 = \boxed{8} \)

\( S_2 = \boxed{24} \)

\( S_3 = \boxed{56} \)

\( S_4 = \boxed{120} \)