Question

P(x) is a function for which P(1) = 1, P(2) = 2, P(3) = 3, and P(x + 1) = [P(x-2) P(x-1) + 1] / P(x) for x >= 3. Find the value of P(6).

I have no idea how do even start and help would be very very appreciated.

Answer 1

Let's try an analyze \(P(4)\). You can use the formula \(P(x+1)\) if you say that \(P(x+1)=P(4)\), implying that \(x=3\)

\[P(4)=P(3+1)=\frac{P(3-2)P(3-1)+1}{P(3)}=\frac{P(1)P(2)+1}{P(3)} =\frac{(1)(2)+1}{3}\]. Find \(P(5)\) and then \(P(6)\) in a similar way

Answer 2

(The reason why you should start at P(4), then find P(5), then P(6), because if you find P(6) first, you will still have to figure out P(4) and P(5) from the formula)

Answer 3

so for P(5) I got 7/4 and for P(6) I got 13/6

which doesn't seem right...

Answer 4

\[\large P(5)=P(4+1)=\frac{P(4-1)P(4-2)+1}{P(4)} \]

Answer 5

P(3) P(2) + 1 / P(4)
(3)(2) + 1 / 4
7/4

So am I correct then? Or am I doing something wrong?

Answer 6

\[\frac{P(3)P(2)+1}{4}=\frac{P(3)P(2)}{4}+\frac{1}{4}\]

Answer 7

So

P(3) P(2) / 4
(3)(2) / 4 = 6/4
6/4 + 1/4 = 7/4

Yes?

Answer 8

\(P(4) = 1\)

Answer 9

\[P(4)=\frac{(1)(2)+1}{3}=\frac{2+1}{3}=\frac{3}{3}=1 \]

Answer 10

But P(3) = 3 and P(2) = 2
so why is the top line (1)(2) + 1?
I'm sorry, I'm not trying to be difficult, I just really don't understand...

Answer 11

I am looking at P(4) now, not P(5).
P(4) = 1, not 4, because of how I wrote it in my first post. And I just simplified the fraction above to show you P(4) =1, and not 4

Answer 12

Ok I think I got it..

Answer 13

:)