It took me a while to solve this!

Probability Level 5

∑ k = 0 m ( 2 n − k k ) ( 2 n − k n ) ⋅ 2 n − 4 k + 1 2 n − 2 k + 1 ⋅ 2 n − 2 k \large \sum_{k=0}^{m} \,\,\large \frac{{2n-k}\choose{k}}{{2n-k}\choose{n}} \cdot \ \frac{2n-4k+1}{2n-2k+1} \cdot \,2^{n-2k}

Given that n n is a positive integer and m m and n n are non-negative integers such that m ≤ n m \,\leq\, n , which of the given option is the closed form expression of given summation above.

( n m + 4 ) ( 2 n − 2 m n − m + 1 ) 2 n − 2 m \frac{{n}\choose{m+4}}{{2n-2m}\choose{n-m+1}}2^{n-2m} ( n + 2 m ) ( 2 n − 2 m + 2 n − m + 1 ) 2 n − 2 m \frac{{n+2}\choose{m}}{{2n-2m+2}\choose{n-m+1}}2^{n-2m} ( n m ) ( 2 n − 2 m n − m ) 2 n − 2 m \frac{{n}\choose{m}}{{2n-2m}\choose{n-m}}2^{n-2m} ( n + 4 m + 7 ) ( 2 n − 2 m + 9 n − m + 6 ) 2 n − 2 m \frac{{n+4}\choose{m+7}}{{2n-2m+9}\choose{n-m+6}}2^{n-2m}

Aditya Sky by Aditya Sky , 21, India

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1 solution

Mark Hennings
May 8, 2016

∑ k = 0 m ( 2 n − k k ) ( 2 n − k n ) 2 n − 4 k + 1 2 n − 2 k + 1 2 n − 2 k = ∑ k = 0 m n ! ( n − k ) ! k ! ( 2 n − 2 k ) ! 2 n − 4 k + 1 2 n − 2 k + 1 2 n − 2 k = ∑ k = 0 m n ! ( n − k ) ! k ! ( 2 n − 2 k ) ! ( 1 − 2 k 2 n − 2 k + 1 ) 2 n − 2 k = ∑ k = 0 m { n ! ( n − k ) ! k ! ( 2 n − 2 k ) ! 2 n − 2 k − k n ! ( n − k ) ! k ! ( 2 n − 2 k + 1 ) ! 2 n − 2 k + 1 } = ∑ k = 0 m n ! ( n − k − 1 ) ! k ! ( 2 n − 2 k − 1 ) ! 2 n − 2 k − 1 − ∑ k = 1 m n ! ( n − k ) ! ( k − 1 ) ! ( 2 n − 2 k + 1 ) ! 2 n − 2 k + 1 = ∑ k = 0 m n ! ( n − k − 1 ) ! k ! ( 2 n − 2 k − 1 ) ! 2 n − 2 k − 1 − ∑ k = 0 m − 1 n ! ( n − k − 1 ) ! k ! ( 2 n − 2 k − 1 ) ! 2 n − 2 k − 1 = n ! ( n − m − 1 ) ! m ! ( 2 n − 2 m − 1 ) ! 2 n − 2 m − 1 = n ! ( n − m ) ! m ! ( 2 n − 2 m ) ! 2 n − 2 m = ( n m ) ( 2 n − 2 m n − m ) 2 n − 2 m \begin{array}{rcl} \displaystyle \sum_{k=0}^m \frac{\binom{2n-k}{k}}{\binom{2n-k}{n}}\, \frac{2n-4k+1}{2n-2k+1}2^{n-2k} & = & \displaystyle \sum_{k=0}^m \frac{n! (n-k)!}{k! (2n-2k)!}\,\frac{2n-4k+1}{2n-2k+1}\,2^{n-2k} \\ & = & \displaystyle \sum_{k=0}^m \frac{n! (n-k)!}{k! (2n-2k)!} \left(1 - \frac{2k}{2n-2k+1}\right)2^{n-2k} \\ & = & \displaystyle \sum_{k=0}^m \left\{ \frac{n! (n-k)!}{k! (2n-2k)!}\,2^{n-2k} - k\,\frac{n!(n-k)!}{k! (2n-2k+1)!}\,2^{n-2k+1}\right\} \\ & = & \displaystyle \sum_{k=0}^m \frac{n! (n-k-1)!}{k! (2n-2k-1)!}\,2^{n-2k-1} - \sum_{k=1}^m \frac{n! (n-k)!}{(k-1)! (2n-2k+1)!}\,2^{n-2k+1} \\ & = & \displaystyle \sum_{k=0}^m \frac{n! (n-k-1)!}{k! (2n-2k-1)!}\,2^{n-2k-1} - \sum_{k=0}^{m-1} \frac{n! (n-k-1)!}{k! (2n-2k-1)!}\,2^{n-2k-1} \\ & = & \displaystyle \frac{n! (n-m-1)!}{m! (2n-2m-1)!}\,2^{n-2m-1} \; = \; \frac{n! (n-m)!}{m! (2n-2m)!}\,2^{n-2m} \\ & = & \displaystyle \frac{\binom{n}{m}}{\binom{2n-2m}{n-m}}2^{n-2m} \end{array}

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Nice solution. Does anyone know if there is a nice combinatorial argument for proving this?

Eli Ross Staff - 5 years, 1 month ago

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I'm also very eager to see such a solution.

Aditya Sky - 5 years, 1 month ago

Well done (+1) :).

Aditya Sky - 5 years, 1 month ago

I did the same thing. I messed up in the beggining. We have to be extremely cautious.

Srikanth Tupurani - 2 years, 3 months ago

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