Question

f(1) = 2 and f(2) = 3, f(n) = f(1) + f(2) + f(n - 1), for n > 2. f(5) = ______

Answer 1

I think it's 10, but I'm not sure.

Answer 2

how do you solve it? bc im lost even how to get the answer :/

Answer 3

f(n) = f(1) + f(2) + f(n-1), right?

f(5) = 2 + 3 + f(4)

Assuming that f(x) = x+1, f(4) would be 5

f(5) = 2 + 3 + 5

f(5) = 10

Answer 4

OH ok thank you so much!

Answer 5

for n>2 , apply the formula
so, let take n=3, f(3) = f(1) +f(2) + f(3-1) = f(1) + 2f(2) = 1+ 2*2= 5
n=4, f(4) = f(1) + f(2) +f(4-1) = f(1)+f(2) + f(3) = 1+2 +5 = 8
n=5 , f(5) = f(1) +f(2) +f(5-1) = f(1)+f(2) + f(4) = 1+2+8 = 11

Answer 6

Does it make sense?

Answer 7

yes thanks

Answer 8

uh @loser66 your arithmetic isn't quite right...

\[f(3) = f(1) + f(2) + f(3-1) = f(1)+ f(2) + f(2) = f(1) + 2f(2) \]\[\qquad = 2+2*3 = 8\]

Answer 9

yes, hahaha... thank you @whpalmer4

Answer 10

so.... now im lost....

Answer 11

guide him, please, @whpalmer4

Answer 12

then \(f(4) = f(1)+f(2) + f(4-1) = 2+3 + 8 = \)

Answer 13

I am not good at teaching. :(

Answer 14

that = 13 so now what?
or is 13 the answer?

Answer 15

so \(f(4) = 13\) but we need to find \(f(5)\)

can you tell me the expression for \(f(5)\) from the formula above?

Answer 16

so then would it be f(5)=f(1)+f(2)+f(5-1)?

Answer 17

exactly! and we know \(f(5-1) = f(4) = 13\)

Answer 18

so then 2+3+13=18 so the answer is 18?

Answer 19

yes, that's my belief...

Answer 20

ok, give me a couple of minutes and ill be able to tell you if that's right or not lol

Answer 21

First 10 values: 2, 3, 8, 13, 18, 23, 28, 33, 38, 43

Answer 22

actually it's just a simple arithmetic sequence after the first 2 values, because we always add 2+3 to the previous term...

Answer 23

oh i see

Answer 24

i submitted 18 and its apparently wrong....

Answer 25

are you positive you have shown us the problem exactly as stated? Those values I gave you are the values of the recursive sequence you described.

Answer 26

you had f(2) = 3 in the problem statement here

corrected sequence is 2, 5, 12, 19, 26, 33, 40, 47, 54, 61

Answer 27

sorry... so the answer is 26?

Answer 28

yes, the terms in my list are f(1), f(2), f(3), f(4), f(5), etc.

\[f(5) = f(1) + f(2) + f(4) = f(1) + f(2) + (f(1)+f(2)+(f(1)+f(2)+f(2))) \]\[\qquad= 2+5+(2+5+(2+5+5)) =26\]

Answer 29

ok thx let me try that one

Answer 30

gets a little confusing typing it like that :-)

maybe better would have been
\[f(5) = f(1)+f(2) + f(4)\]\[f(4) = f(1)+f(2) + f(3)\]\[f(3) = f(1)+f(2)+f(2)\]\[f(2) = 5\]\[f(1) =2\]\[f(5) = 2+5+(2+5+(2+5+5)) = \]

this way does make it apparent that now we have an arithmetic sequence with a common difference of 7 instead of 5, because each additional term adds 2+5=7

Answer 31

ok so im being asked for f(1) = 2 and f(2) = 3, f(n) = f(1) + f(2) + f(n - 1), for n > 2.

f(5) = ______
that would be 18?

Answer 32

is this a different problem? in the screen capture you sent, it was f(2) = 5...

Answer 33

yes it is

Answer 34

Okay, then yes, if you work it out the same way, you get \(f(5) = 18\) just as we did when we were doing the "wrong" problem :-)

Answer 35

lol so it is 18

Answer 36

probably best way for you to write it would be\[f(1) = 2\]\[f(2) = 3\]\[f(3) = f(1)+f(2)+f(2) = 2+3 +3 = 8\]\[f(4) = f(1)+f(2) + f(4-1) = f(1)+f(2) + f(3) = 2+3+8 = 13\]\[f(5) = f(1)+f(2) + f(5-1) = f(1)+f(2) + f(4) = 2+3+13 = 18\]
rather than trying to do it recursively like I did

Answer 37

better middle line:
\[f(3) = f(1)+f(2) + f(3-1) = f(1)+f(2)+f(2) = 2+3+3=8\]

Answer 38

baby steps, baby steps :-)

Answer 39

lol

Answer 40

WHOOOOOO!!!!!!!!! that was the right answer thank you soooooooooooooo much!!

Answer 41

Great!