How do you find nth term rule for $1,3,9,27,...$ ?

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How do you find nth term rule for $1,3,9,27,...$ ?

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Answer 1 · 2017-01-31T19:57:12

The n-th term of this sequence appears to be $3^(n-1)$ , $n >= 1$ .

Explanation:

These are powers of $3$ ordered from $3^0 = 1$ to $3^a$ (for an integer $a >=1$ ). However, the first convenient value for $n$ is $1$ , not $0$ (imagine saying the 0th term of a sequence). Because of that, since the first term is actually $3^0$ , we need to start from the first term ( $n=1$ ) being

$3^(1-1)$ . The next is $3^(2-1)$ , $3^(3-1)$ ... $3^(n-1)$ .

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How do you find nth term rule for $1,3,9,27,...$ ?

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How do you find nth term rule for 1,3,9,27,...1,3,9,27,...?

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How do you find nth term rule for $1,3,9,27,...$ ?