To solve this question, let's analyze and follow the steps needed. The question provides the sum of the first $n$ terms of a sequence, $S_n = 3(2^n - 1)$, and asks us to:

1. Show that the sequence is a geometric sequence (geometric series).
2. Determine the value of $n$ for which the sum exceeds $10^5$.

Let's proceed with each part.

Step 1: Prove the Sequence is Geometric

The sum formula for the first $n$ terms of a geometric series with initial term $a$ and common ratio $r$ is: $S_n = a \frac{r^n - 1}{r - 1}$

Given $S_n = 3(2^n - 1)$, we can compare this with the standard form of the geometric sum formula. Here, it appears that:

• The initial term $a = 3$.
• The common ratio $r = 2$.

So, the series has terms of the form: $a_n = a \cdot r^{n-1} = 3 \cdot 2^{n-1}$

This confirms that it is a geometric series with $a = 3$ and $r = 2$.

Step 2: Find $n$ When $S_n > 10^5$

We need to find the smallest $n$ such that: $S_n = 3(2^n - 1) > 10^5$

1. Divide both sides by 3: $2^n - 1 > \frac{10^5}{3}$

2. Calculate $\frac{10^5}{3} \approx 33333.33$, so we have: $2^n > 33334.33$

3. Now, take the logarithm of both sides (using base 2 for simplicity): $n > \log_2(33334.33)$

4. Calculate $\log_2(33334.33) \approx 15.03$, so $n \approx 16$.

Therefore, the smallest integer $n$ for which $S_n > 10^5$ is $n = 16$.

Final Answer

• The sequence is a geometric series with $a = 3$ and $r = 2$.
• The smallest $n$ such that $S_n > 10^5$ is $n = 16$.

Would you like further clarification on any step?

Here are some additional questions to deepen understanding:

1. How do you determine the general term of a geometric sequence?
2. Why do we use logarithms to solve for $n$ in exponential inequalities?
3. What is the significance of the common ratio in a geometric sequence?
4. How would the solution change if the ratio $r$ were different?
5. Can you derive the formula for the sum of an infinite geometric series?

Tip: For geometric sequences, always check the common ratio by dividing consecutive terms.