Identify the Sequence 1/2 , 1/3 , 2/9 , 4/27

$\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{9}$ , $\frac{4}{27}$

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by $\frac{2}{3}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{2}{3}$

This is the form of a geometric sequence.

$a_{n} = a_{1} r^{n - 1}$

Substitute in the values of $a_{1} = \frac{1}{2}$ and $r = \frac{2}{3}$ .

$a_{n} = \frac{1}{2} {(\frac{2}{3})}^{n - 1}$

Apply the product rule to $\frac{2}{3}$ .

$a_{n} = \frac{1}{2} \cdot \frac{2^{n - 1}}{3^{n - 1}}$

Combine.

$a_{n} = \frac{1 \cdot 2^{n - 1}}{2 \cdot 3^{n - 1}}$

Cancel the common factor of $2^{n - 1}$ and $2$ .

Factor $2$ out of $1 \cdot 2^{n - 1}$ .

$a_{n} = \frac{2 (1 \cdot 2^{n - 2})}{2 \cdot 3^{n - 1}}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 3^{n - 1}$ .

$a_{n} = \frac{2 (1 \cdot 2^{n - 2})}{2 (3^{n - 1})}$

Cancel the common factor.

$a_{n} = \frac{2 (1 \cdot 2^{n - 2})}{2 \cdot 3^{n - 1}}$

Rewrite the expression.

$a_{n} = \frac{1 \cdot 2^{n - 2}}{3^{n - 1}}$

Multiply $2^{n - 2}$ by $1$ .

$a_{n} = \frac{2^{n - 2}}{3^{n - 1}}$

Identify the Sequence 1/2 , 1/3 , 2/9 , 4/27

Related Questions

Support LCMGCF.com by disabling your adblocker.