Help needed (Golden Ratio?)

man man

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

15[(1+52)n+1(152)n+1] \huge{\frac{1}{\sqrt{5}}[(\frac{ 1 + \sqrt{5}}{2})^{n + 1} - (\frac{ 1 - \sqrt{5}}{2})^{n + 1}]} n1 n \geq 1

#Combinatorics #JEE

Note by U Z
6 years, 7 months ago

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Comments

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

Let TnT_n be the number of steps required to climb a staircase with nn steps.

T1=1T_1=1

T2=2T_2=2

For n3n\geq3, as @Michael Mendrin mentioned, we can start with one step, and the rest of it are Tn1T_{n-1} steps. If we start on two steps, the rest of it are Tn2T_{n-2} steps. Now we have a conclusion that

Tn=Tn1+Tn2T_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,...1,2,... instead of the typical 1,1,2,...1,1,2,.... Hence,

Tn=Fn+1T_n=F_{n+1}

Christopher Boo - 6 years, 7 months ago

Well, let's see, let's say there's aa ways to climb up n2n-2 steps, and bb ways to climb up n1n-1 steps, so if we backtrack from the nnth step either 11 or 22 steps, then those would be the number of ways to get to those previous steps. Hence, the number of ways to get to the nnth step is a+ba+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 n2n-2 steps, there are actually 22 ways to get to the nnth step, in fact the 1+11+1 steps to the nnth step is already part of counting the ways up to and through the n1n-1th step. So, it's still a string of honest Fibonacci numbers.

Michael Mendrin - 6 years, 7 months ago

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 - 6 years, 7 months ago

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 - 6 years, 7 months ago

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@Pranshu Gaba @Pranav Arora @Mursalin Habib

and others you too please help

U Z - 6 years, 7 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. :)

Agnishom Chattopadhyay - 6 years, 7 months ago

Where's n in that expression??

Pranjal Jain - 6 years, 7 months ago

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O sorry thank you

U Z - 6 years, 7 months ago

Hint:- Use recurrence relations. Define Tn T_n as the number of ways to climb a staircase with n n steps. Shows its relation with Tn1 T_{n-1} and Tn2 T_{n-2} . You should come up with a familiar relation.

Siddhartha Srivastava - 6 years, 7 months ago

I think it has got something to do with Fibonacci numbers cause the expression gives the n+1n+1 th Fibonacci number.

Aneesh Kundu - 6 years, 7 months ago

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

Ronq Vader - 6 years, 7 months ago
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