Apply the Binomial!

Algebra Level 4

Find the least positive integer 'm' such that

( 2 n n ) 1 / n < m {2n \choose n}^{1/n} < m

for all positive integers n.

Note : Here ( n r ) {n \choose r} represents "n choose r".

This is a part of my set "Beautiful.. It is!!" .


The answer is 4.

Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

Ravi Dwivedi
Jul 9, 2015

Note that ( 2 n n ) < ( 2 n 0 ) + ( 2 n 1 ) + ( 2 n 2 ) + . . . + ( 2 n 2 n ) = ( 1 + 1 ) 2 n = 4 n {2n \choose n} < {2n \choose 0} +{2n \choose 1}+{2n \choose 2}+...+{2n \choose 2n} =(1+1)^{2n}=4^{n}

For n = 5 n=5

( 10 5 ) = 252 > 3 5 {10 \choose 5} = 252>3^{5}

So least positive integer is m = 4 \boxed{m=4}

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Moderator note:

That's a nice question. What got you thinking about it?

Great to see your solutions typed up here. Makes it easier to edit it :)

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