A geometric sequence is defined recursively by $a_{n} = 4 a_{n - 1}$ $a_n= 4a_(n-1)$ , and the first term of the sequence is 0.5, how do you find the explicit formula?

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A geometric sequence is defined recursively by $a_{n} = 4 a_{n - 1}$ $a_n= 4a_(n-1)$ , and the first term of the sequence is 0.5, how do you find the explicit formula?

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Answer 1 · 2018-07-03T20:40:24

$a_{n} = 0.5 {(4)}^{n - 1}$ $a_n= 0.5(4)^(n-1)$

Explanation:

Looking at this we have...
$a_{n} = 4 a_{n - 1}$ $a_n= 4a_(n-1)$

$a_{2} = 4 a_{2 - 1}$ $a_2= 4a_(2-1)$
$a_{2} = 4 a_{1}$ $a_2= 4a_1$
$a_{2} = 4 \cdot 0.5$ $a_2= 4*0.5$
$a_{2} = 2$ $a_2=2$

Since we are told that is a geometric sequence there has to be a $r$ $r$ or a common ratio:
$r = \frac{a_{2}}{a_{1}}$ $r=a_2/a_1$
$r = \frac{2}{.5} = 4$ $r=2/.5=4$

So explicitly written, the formula would be:
$a_{n} = 0.5 {(4)}^{n - 1}$ $a_n= 0.5(4)^(n-1)$

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A geometric sequence is defined recursively by $a_{n} = 4 a_{n - 1}$ $a_n= 4a_(n-1)$ , and the first term of the sequence is 0.5, how do you find the explicit formula?

1 Answer

Explanation:

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A geometric sequence is defined recursively by a_n= 4a_(n-1)an=4an−1a_n= 4a_(n-1), and the first term of the sequence is 0.5, how do you find the explicit formula?

1 Answer

Explanation:

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A geometric sequence is defined recursively by $a_{n} = 4 a_{n - 1}$ $a_n= 4a_(n-1)$ , and the first term of the sequence is 0.5, how do you find the explicit formula?