Question

f(1) = 2 and f(2) = 4, f(n) = f(1) + f(2) + f(n - 1), for n > 2?
find f(5)

Answer 1

so if
\[\large f(1) = 2 ... and \rightarrow f(2) = 4 \]
then
\[\large f(n) = f(1) + f(2) + f(n - 1) \]
\[\large f(5) = f(1) + f(2) + f(5 - 1) \]
\[\large f(5) = \color {blue} {f(1)} + \color {green} {f(2)} + f(4) \]
\[\large f(5) = \color {blue} {2} + \color {green} {4} + f(4) \]

now what is f(4)?
@richard67 ?

Answer 2

n1?

Answer 3

i have no idea, sorry, what subject are u studying at the moment? geometric sequences?
i dont even know if you can work it out from here as you're given the numbers for f1 and f2, but then told that there are different rules for when n is greater than 2...?

Answer 4

Algebra

Answer 5

based on the previous 2, and if there wasnt a rule change after n=2, i'd guess that the function f(x) is
\[\huge f(x) = 2^x\] as
\[\large f(1) = 2^1 = 2\] and
\[\large f(2) = 2^2 = 4\]
...but then there's the rule change for x>2

Answer 6

at a guess, we could try continuing...?

Answer 7

\[\large f(4) = 2^4 = 16...?\] therefore
\[\large f(5) = f(1) + f(2) + f(4) \]
\[\large f(5) = 2 + 4 + 16 \]
\[\large f(5) = 22...? (maybe?) \]

Answer 8

it seems to fit the equation, but it may not be the only solution

Answer 9

example: \[\large f(x) = 2 + 2(x-1) \]
\[\large f(1) = 2 + 2(1-1) = 2 + 2(0) = 2 \]
\[\large f(2) = 2 + 2(2-1) = 2+2(1) = 4\]
therefore:
\[\large f(4) = 2 + 2(4-1) = 2+2(6) = 14\]