anonymous
  • anonymous
MEDALS N BECOMING FAN 1.)Find the first six terms of the sequence. a1 = -3, an = 2 ● an-1 -6, -12, -24, -48, -96, -192 -3, -6, -12, -24, -48, -96 0, 2, -6, -4, -2, 0 -3, -6, -4, -2, 0, 2 2.)Find an explicit rule for the nth term of the sequence. 2, -8, 32, -128, ... an = 2 ● 4^(n+1) an = 2 ● (-4)^n an = 2 ● 4^(n-1 ) an = 2 ● (-4)^(n-1)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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DanJS
  • DanJS
The first sequence, is taking the term before it and multiplying it by 2 every time
DanJS
  • DanJS
starting with -3
DanJS
  • DanJS
\[a _{2}=2 * a_{2-1}\] a2 would be 2 times a1 a3 would be 2 times a2 a4 would be 2 times a3 . . .

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More answers

anonymous
  • anonymous
-6, -12, -24, -48, -96, -192
SolomonZelman
  • SolomonZelman
but your first term is?
anonymous
  • anonymous
#2 i got an = 2 ● 4^(n-1 )?
anonymous
  • anonymous
-3, -6, -12, -24, -48, -96
anonymous
  • anonymous
sry i meant that 1
SolomonZelman
  • SolomonZelman
yeah that is better
SolomonZelman
  • SolomonZelman
and for 2 u got?
anonymous
  • anonymous
is my answer for 2 correct?
SolomonZelman
  • SolomonZelman
you are multiplying times what every time?
DanJS
  • DanJS
Since they gave you multiple choice, you can try each one and really know nothing..
SolomonZelman
  • SolomonZelman
(the form is a\(_{\rm n}\)=(a\(_{\rm 1}\))•r\(^{\rm n-1}\) ) r - common ratio - (the number by which you multiply each time)
SolomonZelman
  • SolomonZelman
in this case: 2, -8, 32, -128 you are multiplying times 4? That would be close, but not exactly right, because your values are changing signs (they alternate)
SolomonZelman
  • SolomonZelman
2 • what = -8? -8• what = 32
anonymous
  • anonymous
an = 2 ● (-4)^n @SolomonZelman
SolomonZelman
  • SolomonZelman
ok, i am not going to tell you right or wrong, but this is what I will say \(a_1=2\cdot (-4)^{1}\) \(a_1=2\cdot(-4)\) \(a_1=-8\)
SolomonZelman
  • SolomonZelman
but -8 is the second term, not the first
anonymous
  • anonymous
oh so it has to be this one an = 2 ● (-4)^(n-1)
SolomonZelman
  • SolomonZelman
now, \(a_n=2\cdot(-4)^{n-1}\) \(a_2=2\cdot(-4)^{2-1}=2\cdot (-4)^1=-8\) AND \(a_1=2\cdot(-4)^{1-1}=2\cdot (-4)^0=2 \cdot 1 = 2\). ---------------------------------- YES IT HAS TO BE \(a_n=2\cdot(-4)^{n-1}\)
SolomonZelman
  • SolomonZelman
u r done with this question:)
anonymous
  • anonymous
thank u!
SolomonZelman
  • SolomonZelman
Anytime!

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