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What is the rule for the following geometric sequence? 64,128,256,512...
a) a_(n) = 64(-2)^n-1
b) a_(n) = 64(2)^n-1
c) a_(n) = 64(1/2)^n-1
d) a_(n) = 64(-1/2)^n-1

Sagot :

Answer:

B

Step-by-step explanation:

The explicit rule for a geometric sequence is

[tex]a_{n}[/tex] = a₁ [tex](r)^{n-1}[/tex]

where a₁ is the first term and r the common ratio

Here a₁ = 64 and r = [tex]\frac{a_{2} }{a_{1} }[/tex] = [tex]\frac{128}{64}[/tex] = 2 , then

[tex]a_{n}[/tex] = 64 [tex](2)^{n-1}[/tex] → B

Answer:

b).

[tex]{ \tt{a _{n} = a( {r}^{n - 1} )}} \\ { \tt{a _{n} = 64( {2}^{n - 1}) }}[/tex]

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