The given series is:

$32, 8, 2, \dots$

This is a geometric series because each term is obtained by multiplying the previous term by a common ratio $r$. Let's determine the common ratio:

$r = \frac{8}{32} = \frac{1}{4}$

So, the common ratio is $\frac{1}{4}$. The first term $a$ is $32$.

Formula for the sum of the first $n$ terms of a geometric series:

$S_n = \frac{a(1 - r^n)}{1 - r}$ where:

• $a$ is the first term,
• $r$ is the common ratio,
• $n$ is the number of terms.

For the first 9 terms:

• $a = 32$,
• $r = \frac{1}{4}$,
• $n = 9$.

Substitute the values into the formula:

$S_9 = \frac{32(1 - (\frac{1}{4})^9)}{1 - \frac{1}{4}}$

Simplifying the denominator:

$1 - \frac{1}{4} = \frac{3}{4}$

So, the sum becomes:

$S_9 = \frac{32(1 - (\frac{1}{4})^9)}{\frac{3}{4}} = \frac{32 \times 4}{3} \times \left(1 - \frac{1}{262144}\right)$

$S_9 = \frac{128}{3} \times \left(1 - 0.0000038\right)$

$S_9 \approx \frac{128}{3} \times 0.9999962 \approx 42.6667 \times 0.9999962 \approx 42.6666$

Thus, the sum of the first 9 terms of the series, to the nearest integer, is:

$\boxed{43}$

Would you like more details on any step or have any further questions?

Related questions:

1. How do you find the common ratio of a geometric series?
2. How would the formula for the sum change if the common ratio were negative?
3. What is the sum of the first 5 terms of this series?
4. How do you determine if a series is geometric?
5. Can the sum formula be applied to infinite geometric series?

Tip:

In a geometric series, the sum converges only if the absolute value of the common ratio $|r| < 1$.