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? (4 points) -5, -20, -65, -200, -5, 20, -92, 372, -5, -24, -92, -372, -5, 20, -65, 200,
@DanJS @zepdrix
@3mar
\[\large\rm f(\color{orangered}{n})=-4\cdot f(\color{orangered}{n}-1)-3\cdot f(\color{orangered}{n}-2)\] So umm.. you should be able to eliminate one of your options immediately. If our first term is -5, and the next is 20, you can ignore the list that doesn't start with -5 and 20.
Plug in n=2 to find the third term of your sequence,\[\large\rm f(\color{orangered}{2})=-4\cdot f(\color{orangered}{2}-1)-3\cdot f(\color{orangered}{2}-2)\]\[\large\rm f(\color{orangered}{2})=-4\cdot f(1)-3\cdot f(0)\]
They gave you the values for f(0) and f(1). So plug them in and simplify.
whaaa
this is so confusing
This is so easy! The first term is -5 The second term is 20 Any choice does not match these two criteria is excluded! Is it clear so far?
what other choices are there?
-5, -20, -65, -200..........× -5, 20, -92, 372.............may be -5, -24, -92, -372...........× -5, 20, -65, 200.............may be
oh okay so we can eliminate C
and A
Congratulations!
@zepdrix Please proceed!
Where are you guys?
oh sorry im back
So Are you satisfied?
yes
he wont come back i think so you will have to do next steps
@3mar can we work on this together quickly
Pleasure is mine!
Yes, can you?
yes wait 2 min
please thanks
As @zepdrix said and mentioned: \[\large\rm f(\color{orangered}{2})=-4\cdot f(\color{orangered}{2}-1)-3\cdot f(\color{orangered}{2}-2)\] \[\large\rm f(\color{orangered}{2})=-4\cdot f(1)-3\cdot f(0)\]
Ca you find out the third term!?
@3mar sooo sorry
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