The table below shows the values of f(n) for different values of n: n 1 2 3 4 5 6 f(n) 1 2 5 12 29 70 Which recursive function best represents the values shown in the table? f(1) = 1, f(2) = 2, f(n) = 2f(n -1) f(n - 2); n > 2 f(1) = 1, f(2) = 2, f(n) = f(n -3) + f(n - 2); n > 2 f(1) = 1, f(2) = 2, f(n) = 2f(n -1) + f(n - 2); n > 2 f(1) = 1, f(2) = 2, f(n) = f(n -3) f(n - 2); n > 2
@ganeshie8 plz help
do you know how to do a sequence? @hero a lil help
well i know that we replace the number oh wait... Arithmetic Recursive Formula Examine the recursive formula given the arithmetic sequence 2, 5, 8, 11, … The first term of the sequence is 2, and the common difference is 3. f(1) = 2, d = 3 To find the next term in the sequence, the common difference is added to the previous term. f(2) = f(1) + d = 2 + 3 = 5 f(3) = f(2) + d = 5 + 3 = 8 To find the nth term of this sequence, the common difference would be added to the term before it. This is the generic form for an arithmetic recursive formula. f(n) = f(n – 1) + d The variable d is the common difference and n > 0. Check to see that this works by finding the fourth term of the sequence. (n = 4)
is it a @muzzack ?
@ganeshie8 plz help
well what do you think??
no idea lol
well believe ur gut
which rule fits?
idk lol
so the question becomes .. why dont you know? you have 4 options to test out, and they give you the values ....
what are the values? what are you supposed to do with the options? (you can see im not really good at this stuff)
the values are listed for you already ... for exmaple, when n=3, f(3)=5 you have to determine which rule best defines the function.
notice that there is no n=0 .. so f(n-3) means nothing to us when n=3. what does this tell us about options with f(n-3) in them?
tht f(0)=0 i think?
f(0) is not defined.
i meant tht lol
so i have to do f(?)
so, we can rule out any option that has f(n-3) in it since they all say n starts at 3, and f(3-3) is simply undefined.
So there is only a and c left over
right, so now lets follow the rules defined: f(n) = 2f(n -1) f(n - 2); n > 2 , so let n=3 f(3) = 2f(3 -1) f(3 - 2) f(3) = 2f(2) f(1) , is this true? --------------------------------- f(n) = 2f(n -1) + f(n - 2) f(3) = 2f(3 -1) + f(3 - 2) f(3) = 2f(2) + f(1), or is this true?
hold on i have to substitute the values in the n right? give me a few mins to do tht
yes
The second one right @amistre64 ?
yes: 5 = 2(2)+1 is true.
Thank You!
now lets see if it works for the next value, im sure it does, 12 = 5(2) + 2 ... seems to work fine
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