anonymous
  • anonymous
Find the first, fourth, and eighth terms of the sequence. A(n) = -2 x 2^x-1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
there is no "n" in your formula
anonymous
  • anonymous
i got 1; -64; -16,384
anonymous
  • anonymous
probably wrong no idea

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anonymous
  • anonymous
you have \(A_n\) on the left, but no \(n\) on the right maybe a typo?
anonymous
  • anonymous
\[A_n=-2\times 2^{n-1}\] perhaps?
anonymous
  • anonymous
|dw:1370225426801:dw|
anonymous
  • anonymous
sorry yeah
anonymous
  • anonymous
ok then if \(n=1\) you get \[A_1=-2\times 2^{1-1}=-2\times 2^0=-2\times 1=-2\]
anonymous
  • anonymous
if \(n=2\) you get \[A_2=-2\times 2^{2-1}=-2\times 2^1=-4\] etc
anonymous
  • anonymous
so -4; -32; -512?
anonymous
  • anonymous
or no -2; -16; -256
Loser66
  • Loser66
you have to calculate by yourself, no one can check your solution . if you want to make sure about your answer, you should construct A_n by recursive such that A_n depends on A1 only. At that time, no matter how large n is, you can calculate it without depending on the previous term.
Loser66
  • Loser66
this problem asks you find A_8 only, quite easy to calculate 7 times before getting A_8. what if they ask you find A_100? calculate 99 times?

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