The given sequence is $128, 64, 32, 16, \ldots$, which is a geometric sequence because each term is obtained by multiplying the previous term by a constant ratio.

Step 1: Identify the first term and the common ratio

• The first term ($a$) is $128$.
• The common ratio ($r$) is obtained by dividing any term by the previous term: $r = \frac{64}{128} = \frac{1}{2}.$

Step 2: Formula for the sum of the first $n$ terms of a geometric sequence

The formula for the sum of the first $n$ terms of a geometric sequence is: $S_n = a \frac{1 - r^n}{1 - r}, \quad \text{if } r \neq 1.$

Step 3: Substitute the values

• $a = 128$
• $r = \frac{1}{2}$
• $n = 13$

$S_{13} = 128 \frac{1 - \left(\frac{1}{2}\right)^{13}}{1 - \frac{1}{2}}$

Step 4: Simplify the denominator

$1 - \frac{1}{2} = \frac{1}{2}.$

$S_{13} = 128 \cdot 2 \cdot \left(1 - \left(\frac{1}{2}\right)^{13}\right)$

Step 5: Calculate $\left(\frac{1}{2}\right)^{13}$

$\left(\frac{1}{2}\right)^{13} = \frac{1}{8192} \approx 0.000122.$

$S_{13} = 128 \cdot 2 \cdot \left(1 - 0.000122\right)$

$S_{13} = 128 \cdot 2 \cdot 0.999878$

$S_{13} \approx 128 \cdot 1.999756 = 255.97.$

Final Answer:

The sum of the first 13 terms is approximately 255.97.

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Related Questions:

1. What is the sum of the first 20 terms of this sequence?
2. How does the sum behave as $n \to \infty$?
3. Can this formula be applied to sequences with $r > 1$?
4. How would the sum change if the sequence started at $64$ instead of $128$?
5. What is the general term ($a_n$) of this sequence?

Tip:

For geometric sequences, if $|r| < 1$, the sum of infinitely many terms converges to $\frac{a}{1-r}$.