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RMO 2013 problem

Could someone please help me with the attached RMO 2013 problem?

I got the answer to be \(3^{n-1}n + [\frac{n-1}{2}]\), where the square brackets, [], imply the floor function. Is that right? I couldn't find an answer key for this (Maharashtra and Goa) anywhere.

Any help will be highly appreciated.

Note by Shashank Rammoorthy
1 year, 11 months ago

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I'm getting \( 2^{n-1}(2^n - 1) \).

Siddhartha Srivastava - 1 year, 11 months ago

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Could you please explain?

Shashank Rammoorthy - 1 year, 11 months ago

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Let \( b_n \) be the number of sequences formed which have an even number of zeros.

Then it is obvious that \( 4^n = a_n + b_n \), since there \( 4^n \) sequences of \( n \) terms, and each sequence must contain an even or odd number of zeroes.

Now, look at any sequence part of \( a_n \). It's first term is either zero (1 way), or not zero (3 ways). If the first term is zero, the remaining sequence must contain an even number of zeroes, and thus is \( b_{n-1} \). Likewise, if the first term is not zero, the remaining sequence is \( a_{n-1} \).

Therefore, \( a_n = b_{n-1} + 3a{n-1} \). Eliminating \( b _n \) from both equations, we have

\( a_n = 2a_{n-1} + 4^{n-1} \)

or \( a_n = 6a_{n-1} -8a_{n-2} \)

Solving this recurrence relation, using \( a_1 = 1, a_2 = 6, \), we get \( a_n = 2^{n-1}(2^n - 1) \)

Siddhartha Srivastava - 1 year, 11 months ago

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@Siddhartha Srivastava Thanks!

Shashank Rammoorthy - 1 year, 11 months ago

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