Identify the Sequence 8 , 32 , 128

$8$ , $32$ , $128$

Step 1

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 $4$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = 4$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 8$ and $r = 4$ .

$a_{n} = 8 \cdot 4^{n - 1}$

Step 4

Multiply $8 \cdot 4^{n - 1}$ .

Rewrite $8$ as $2^{3}$ .

$a_{n} = 2^{3} \cdot 4^{n - 1}$

Rewrite $4$ as $2^{2}$ .

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

Multiply the exponents in ${(2^{2})}^{n - 1}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$a_{n} = 2^{3 + 2 n - 2}$

Subtract $2$ from $3$ .

$a_{n} = 2^{2 n + 1}$

Identify the Sequence 8 , 32 , 128

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