A staircase has n steps. A man climbs either one step or two steps at a time .Prove that the number of ways in which he can climb up the staircase , starting from the bottom , is

\( \huge{\frac{1}{\sqrt{5}}[(\frac{ 1 + \sqrt{5}}{2})^{n + 1} - (\frac{ 1 - \sqrt{5}}{2})^{n + 1}]}\) \( n \geq 1 \)

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TopNewestThe answer is already mentioned, so I'll just express it algebraically.

Let \(T_n\) be the number of steps required to climb a staircase with \(n\) steps.

\(T_1=1\)

\(T_2=2\)

For \(n\geq3\), as @Michael Mendrin mentioned, we can start with one step, and the rest of it are \(T_{n-1}\) steps. If we start on two steps, the rest of it are \(T_{n-2}\) steps. Now we have a conclusion that

\(T_n=T_{n-1}+T_{n-2}\)

This is actually the Fibonacci Sequence, but we have to shift it to the right, since it starts with \(1,2,...\) instead of the typical \(1,1,2,...\). Hence,

\(T_n=F_{n+1}\) – Christopher Boo · 2 years, 4 months ago

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@Christopher Boo @Michael Mendrin @Aneesh Kundu @Siddhartha Srivastava @Agnishom Chattopadhyay – Megh Choksi · 2 years, 4 months ago

Thank you sirLog in to reply

The number of ways is just the fibonacci sequence. You can see this by noting the number of ways to climb n steps is equal to the number of ways where the second to last step gets you up n-2 steps, plus the number of ways where the second to last step gets you up n-1 steps. If s(n) represents the number of ways to climb up n steps, then it follows that s(n) + s(n+1) = s(n+2). Since s(1) = 1 and s(2) = 2 by inspection, this is just the fibonacci sequence shifted by 1, as expressed in the formula. – Rogers Epstein · 2 years, 4 months ago

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Well, let's see, let's say there's \(a\) ways to climb up \(n-2\) steps, and \(b\) ways to climb up \(n-1\) steps, so if we backtrack from the \(n\)th step either \(1\) or \(2\) steps, then those would be the number of ways to get to those previous steps. Hence, the number of ways to get to the \(n\)th step is \(a+b\), and we're looking at Fibonacci numbers. The formula give was a dead giveaway to the answer to this one. Thanks for the hint!

Addendum: If one wonders if starting at \(n-2\) steps, there are actually \(2\) ways to get to the \(n\)th step, in fact the \(1+1\) steps to the \(n\)th step is already part of counting the ways up to and through the \(n-1\)th step. So, it's still a string of honest Fibonacci numbers. – Michael Mendrin · 2 years, 4 months ago

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We can notice that this looks similar to Binet’s Theorem, which is an explicit formula to give the nth number in the fibonacci sequence. In this case, it gives the (n+1)th term. The Fibonacci sequence is 1 1 2 3 5 8… while this sequence is 1 2 3 5 8… because the number of ways to get to each step adds up. This sequence is just the Fibonacci sequence with the first two terms as 1 and 2. QED – Jonathan Yang · 2 years, 4 months ago

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I encountered a similar question in the NMTC screening test this year. – Ronq Vader · 2 years, 4 months ago

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I think it has got something to do with Fibonacci numbers cause the expression gives the \(n+1\) th Fibonacci number. – Aneesh Kundu · 2 years, 4 months ago

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Hint:- Use recurrence relations. Define \( T_n \) as the number of ways to climb a staircase with \( n \) steps. Shows its relation with \( T_{n-1} \) and \( T_{n-2} \). You should come up with a familiar relation. – Siddhartha Srivastava · 2 years, 4 months ago

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Where's n in that expression?? – Pranjal Jain · 2 years, 4 months ago

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– Megh Choksi · 2 years, 4 months ago

O sorry thank youLog in to reply

@DEEPANSHU GUPTA @Sanjeet Raria @Sandeep Bhardwaj @Calvin Lin @Michael Mendrin @Pranjal Jain @Agnishom Chattopadhyay @Aditya Raut @Krishna Sharma

Please help – Megh Choksi · 2 years, 4 months ago

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– Agnishom Chattopadhyay · 2 years, 4 months ago

As Christopher and Michael pointed out, you should have tried generating the first few terms of the series. You'll find it's a fibonacci series and if you are not familiar with the formula, look it up. :)Log in to reply

@Pranshu Gaba @Pranav Arora @Mursalin Habib

and others you too please help – Megh Choksi · 2 years, 4 months ago

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