How do you rewrite each explicit formula in function form $g_{n} = - 6 \cdot {(\frac{1}{3})}^{n - 1}$ $g_n=-6*(1/3)^(n-1)$ ?

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Answer 1 · 2018-01-13T11:25:04

The geometric sequence is $-6 , -2 , -2/3 , -2/9 ....$

$g_n = -6 (1/3)^(n-1)$

$g_1 = -6 (1/3)^(1-1) = -6*1= -6$

$g_2 = -6 (1/3)^(2-1) = -6* 1/3= -2$

$g_3= -6 (1/3)^(3-1) = -6* (1/3)^2=-2/3$

$g_4= -6 (1/3)^(4-1) = -6* (1/3)^3=-2/9$

The geometric sequence is $-6 , -2 , -2/3 , -2/9 ....$

First term is $g_1= -6$ , common ratio is $r= -1/3$ [Ans]

How do you rewrite each explicit formula in function form g_n=-6*(1/3)^(n-1)gn=−6⋅(13)n−1g_n=-6*(1/3)^(n-1)?