anonymous
  • anonymous
The following function defines a recursive sequence. f(0) = -5 f(1) = 20 f(n) = -4•f(n -1) - 3•f(n - 2); for n > 1 Which of the following sequences is defined by this recursive function? -5, -20, -65, -200, … -5, 20, -92, 372, … -5, -24, -92, -372, … -5, 20, -65, 200, …
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
what is the first term?
anonymous
  • anonymous
f(0) = -5 ?
amistre64
  • amistre64
yes, so they give you the first one what is the second term?

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amistre64
  • amistre64
we can narrow the options by eliminating the ones that have the wrong first and second terms, then the third term gets us to only one option that fits
amistre64
  • amistre64
we can use the rule given to determine f(2) f(n) = -4•f(n -1) - 3•f(n - 2) f(2) = -4•f(2 -1) - 3•f(2 - 2) f(2) = -4•f(1) - 3•f(0) ^^ ^^ we already know these values to plug into the rule
anonymous
  • anonymous
ok so what now ?
amistre64
  • amistre64
that IS the "what now" you work it out
amistre64
  • amistre64
or you ask questions about what it is you dont understand
anonymous
  • anonymous
oh ok hold on
anonymous
  • anonymous
i dont get what im suppose to work out ....
amistre64
  • amistre64
f(n) = -4•f(n -1) - 3•f(n - 2) f(2) = -4•f(2 -1) - 3•f(2 - 2) f(2) = -4•f(1) - 3•f(0) ^^ ^^ we already know these values to plug into the rule
amistre64
  • amistre64
f(0) = -5 f(1) = 20
amistre64
  • amistre64
wherever you see f(0), replace it by -5 wherever you see f(1), replace it by 20
anonymous
  • anonymous
what if i see f(2)
amistre64
  • amistre64
f(2) is what we are calculating ... f(2) will equal ______________
anonymous
  • anonymous
f(2) = -4•f(1) - 3•f(0) =-65 correct?
amistre64
  • amistre64
very good :)
amistre64
  • amistre64
so what we are looking for is a sequence that starts out: f(0), f(1), f(2) -5, 20, -65 only one of the options starts out like this
anonymous
  • anonymous
the last one ..
amistre64
  • amistre64
correct
anonymous
  • anonymous
thank you very much for explaining
amistre64
  • amistre64
good luck :)

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