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(2n!)/(n!*n!*2^(n+1)) Is this an integer or not ?
asked in Mathematical Logic by (151 points) 1 4 | 52 views
not.

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We may use heat and trial method to solve. 

For N = 1 , denomitor becomes larger then numerator

Same for N = 2.

Hence it's result is not an integer.

May be for larger N it divides, becuause for large N upper part , will give large result 

If we multiply number and denomination by ( N + 1 )  then it becomes  : ( (2 * N ) !  * / ( N !  * ( N + 1 ) ! )) *( (N +1 )  /2 ^(N + 1 ) )  

where first term : ( (2 * N ) !  * / ( N !  * ( N + 1 ) ! )) is catalan number : https://en.wikipedia.org/wiki/Catalan_number 

Hence it will reduces to Nth Catalan number  * ( (N+1) / 2^(N+1) )   where catalan numbers will grow with large factor  as N will increase as compared to power of 2 , and it may divide and result will be integer for some N

But overall if for any N it can't divide then answer will be not !

This type of problem are mostly based on observation, it's not necessary to do proof and all , as in gate exam speed matters, i will go for heat and trial first ( it saves time ) 

answered by Loyal (2.9k points) 3 17
selected by
This is false for every value of n. In all the cases there is always exactly one extra 2 and nothing else which is left in denominator after cancelling all common terms. This can be easily observed by breaking both numerator and denominator into powers of prime factors.

It is a fact that if any number whose denominator contains power of any prime number (after breaking both into powers of prime factors) extra than its corresponding power  in numerator, then that quantity can never be an integer.

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