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this is the question

given $n$ stairs, how many number of ways can you climb either step up one stair or hop up two? I need to include the number of ways for $n=1$ through $6$ as well.

My question : is step up two stairs and hop up two, is it the same thing? however, I tried to do the solution

$n=1$ stair No. of ways $ = 1 = f(2) $

$n= 2$ No. of ways $ = 2 = f(3) = f(2) + f(1) $

$n= 3$ No. of ways $ = 3 = f(4) = f(3) + f(2) $

$n= 4$ No. of ways $ = 5 = f(5) = f(4) + f(3) $

$n= 5$ No. of ways $ = 8 = f(6) = f(5) + f(4) $

$n= 6$ No. of ways $ = 13 =f(7) = f(6) + f(5) $

:

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for $n$, No. of ways $ = f(n+1) $

because there is a hint saying I could use recursively defined.

Is there other way?

Mohsen Shahriari
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guest11
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    but there is nothing to be formulated? – guest11 Dec 01 '15 at 05:43
  • See possibly http://math.stackexchange.com/questions/789804/how-many-distinct-ways-to-climb-stairs-in-1-or-2-steps-at-a-time?rq=1 – turkeyhundt Dec 01 '15 at 05:45
  • thank you :) but do you know why do we use FIbonacci numbers? – guest11 Dec 01 '15 at 05:53
  • All solutions in the link uses fibonacci as well, and I would say there isn't really another "elegant" way as the closed formula involves $\sqrt{5}$. But you can always count all cases and add them up, while this will not give you a closed formula. – cr001 Dec 01 '15 at 05:53
  • There is also a combinatorial solution: http://ms.appliedprobability.org/data/files/Articles%2047/47-1-4.pdf "The Staircase Problem and Fibonacci Numbers" by SIMON ZASLAVSKY and ROGER KHAZAN in Applied Probability Trust 2014 – PT272 Jul 07 '16 at 12:35

1 Answers1

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You have correctly observed your recursive pattern. Namely the Fibonacci Numbers defined by:

$$F(n+2)=F(n+1)+F(n),F(0)=1,F(1)=1$$ (Note different definitions define different start conditions, this is to fit into your situation)

Now it is possible to derive a formula who gives $F(n)$ only in terms of $n$ but the method is significantly harder than the recursive method above.

It is given by:

$$F(n)=\frac{\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n}{\sqrt5}$$

Note this one starts from 0,1 for the first two terms which is slightly different to your answer which has a zeroth term and starts 1,1.

Given that you only need to do up to $n=6$ this formula is a huge overkill and highly likely beyond what your teacher expected you to do.

Ian Miller
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