We are given the sequence $\{a_n\}$ defined as follows:

$a_1 = 2, \quad a_{n+1} = 2(a_1 + a_2 + \cdots + a_n) \quad \text{for} \quad n = 1, 2, 3, \dots$

Part (i): Evaluate the numerical values of $a_2$, $a_3$, and $a_4$

We will compute the first few terms of the sequence.

• $a_1 = 2$
• To find $a_2$, we use the formula: $a_2 = 2(a_1) = 2(2) = 4$
• To find $a_3$, we use: $a_3 = 2(a_1 + a_2) = 2(2 + 4) = 2(6) = 12$
• To find $a_4$, we use: $a_4 = 2(a_1 + a_2 + a_3) = 2(2 + 4 + 12) = 2(18) = 36$

Thus, the first four terms of the sequence are: $a_1 = 2, \quad a_2 = 4, \quad a_3 = 12, \quad a_4 = 36$

Part (ii): Find the ratio $\frac{a_{n+1}}{a_n}$ and determine if the sequence is geometric

To find the ratio $\frac{a_{n+1}}{a_n}$ for $n \geq 2$, we calculate the ratios for specific values of $n$.

• For $n = 2$: $\frac{a_3}{a_2} = \frac{12}{4} = 3$
• For $n = 3$: $\frac{a_4}{a_3} = \frac{36}{12} = 3$

In both cases, the ratio $\frac{a_{n+1}}{a_n} = 3$. This suggests that the ratio is constant for $n \geq 2$.

Is $\{a_n\}$ a geometric sequence?

A sequence is geometric if the ratio between consecutive terms is constant. Since we have found that the ratio $\frac{a_{n+1}}{a_n} = 3$ for $n \geq 2$, the sequence $\{a_n\}$ for $n \geq 2$ is indeed a geometric sequence with common ratio 3.

Summary:

• (i) The numerical values are: $a_2 = 4, \quad a_3 = 12, \quad a_4 = 36$
• (ii) The ratio $\frac{a_{n+1}}{a_n} = 3$ for $n \geq 2$, and the sequence $\{a_n\}$ for $n \geq 2$ is a geometric sequence with common ratio 3.

Do you have any questions or want more details?

Here are 5 related questions to expand your understanding:

1. What would $a_5$ be in this sequence?
2. Can you express $a_n$ in terms of $n$ without using recursion?
3. What is the general form of a geometric sequence?
4. How would the sequence change if the multiplier was different, say 3 instead of 2?
5. Could this sequence represent any real-world scenario, such as in finance or physics?

Tip: To verify if a sequence is geometric, always check if the ratio $\frac{a_{n+1}}{a_n}$ is constant across all $n$.