Question

Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{u_{n}\right\\}=\left\\{\frac{2^{n}}{3^{n-1}}\right\\} $$

Short answer

The sequence is geometric with a common ratio of \( \frac{2}{3} \). The first four terms are 2, \( \frac{4}{3} \), \( \frac{8}{9} \), and \( \frac{16}{27} \).

Step-by-step solution

Identify the General Term

Examine the given sequence that is defined by the general term \( u_n = \frac{2^n}{3^{n-1}} \).

Confirm It's a Geometric Sequence

To confirm if a sequence is geometric, the ratio of successive terms should be constant. Calculate the ratio \( r = \frac{u_{n+1}}{u_n} \).

Calculate the Ratio

Compute \( u_{n+1} \) from the given formula: \( u_{n+1} = \frac{2^{n+1}}{3^n} \).Next, find \( r = \frac{u_{n+1}}{u_n} = \frac{\frac{2^{n+1}}{3^n}}{\frac{2^n}{3^{n-1}}} \). Simplify this to get \( r = \frac{2 \times 2^n}{3 \times 3^{n-1}} \times \frac{3^{n-1}}{2^n} = \frac{2}{3} \).

Write the Common Ratio

The common ratio \( r \) of the geometric sequence is \( \frac{2}{3} \).

List the First Four Terms

Substitute \( n = 1, 2, 3, 4 \) into the general term:- \( u_1 = \frac{2^1}{3^0} = 2 \)- \( u_2 = \frac{2^2}{3^1} = \frac{4}{3} \)- \( u_3 = \frac{2^3}{3^2} = \frac{8}{9} \)- \( u_4 = \frac{2^4}{3^3} = \frac{16}{27} \)

Key concepts

Common Ratio

When discussing geometric sequences, one of the key elements is understanding the common ratio. The common ratio is a constant value between successive terms of a geometric sequence.
This means that if you take any term in the sequence and divide it by the previous term, you get the same value each time. In mathematical terms, if the sequence is given by, u_n, the common ratio, r, is found by:
\( r = \frac{u_{n+1}}{u_n} \).
In our exercise, the general term of the sequence is \(u_n = \frac{2^n}{3^{n-1}} \).
To find the common ratio, substitute \( n+1 \) into the general term:
\( u_{n+1} = \frac{2^{n+1}}{3^n} \).
Then, the ratio is computed as:
\[ r = \frac{u_{n+1}}{u_n} = \frac{\frac{2^{n+1}}{3^n}}{\frac{2^n}{3^{n-1}}} = \frac{2 \times 2^n}{3 \times 3^{n-1}} \times \frac{3^{n-1}}{2^n} = \frac{2}{3} \].
This calculation shows that the common ratio \( r \) for this sequence is \( \frac{2}{3} \).

Sequence Terms

Understanding the terms of a sequence helps unravel its pattern. The terms in a geometric sequence follow a specific repetitive multiplication rule.
Given the general term \( \frac{2^n}{3^{n-1}} \), we can calculate specific terms by plugging in values of n.
Let's list the first four terms:

• For \( n = 1 \), \( u_1 = \frac{2^1}{3^0} = 2 \)


• For \( n = 2 \), \( u_2 = \frac{2^2}{3^1} = \frac{4}{3} \)


• For \( n = 3 \), \( u_3 = \frac{2^3}{3^2} = \frac{8}{9} \)


• For \( n = 4 \), \( u_4 = \frac{2^4}{3^3} = \frac{16}{27} \)

Notice how each term is formed by raising 2 to increasing powers while also dividing by increasing powers of 3. This pattern is a characteristic feature of geometric sequences.

Geometric Progression

A geometric progression is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio.
This property makes geometric sequences easy to identify and work with.
To confirm that a sequence is geometric, ensure that the ratio of successive terms remains consistent. In our sequence \( u_n = \frac{2^n}{3^{n-1}} \), we have already calculated the common ratio as \( \frac{2}{3} \).
This tells us that any term can be obtained by multiplying the previous term by \( \frac{2}{3} \).
The terms \( 2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27} \) showcase this property. If you divide any term by its preceding term, you get \( \frac{2}{3} \), confirming our sequence as geometric.

• This multiplication rule (using the common ratio) keeps the sequence progressing in a definable manner.
• This progression can be predictably extended by continuing to multiply by the common ratio.

By understanding this, you can handle various geometric sequences with confidence.