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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## Comments

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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,

$T_n=F_{n+1}$

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Thank you sir @Christopher Boo @Michael Mendrin @Aneesh Kundu @Siddhartha Srivastava @Agnishom Chattopadhyay

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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.

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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.

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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

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

Please help

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@Pranshu Gaba @Pranav Arora @Mursalin Habib

and others you too please help

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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. :)

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Where's n in that expression??

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O sorry thank you

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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.

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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.

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I encountered a similar question in the NMTC screening test this year.

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