Binomial coefficient inequality equation proofs Prove the statement: If n is a positive integer, t

Cierra Castillo 2022-06-29 Answered
Binomial coefficient inequality equation proofs
Prove the statement:
If n is a positive integer, then 1 = ( n 0 ) < ( n 1 ) < . . . < ( n n / 2 ) = ( n n / 2 ) > . . . > ( n n 1 ) > ( n n ) = 1
I tried to read up the falling/rising factorial notation from internet sources, but I cannot find how to write up the first step.
The options are as follows (We have to put it in the right order, one option is incorrect):
- If k <= n / 2, then k / ( n k + 1 ), so the "greater than" signs are correct. Similarly, if k > n / 2, then k / ( n k + 1 ), so the "less than" signs are correct.
- If k <= n / 2, then k / ( n k + 1 ), so the "less than" signs are correct. Similarly, if k > n / 2, then k / ( n k + 1 ), so the "greater than" signs are correct.
- Since n / 2 + n / 2 = n, the equalities at the ends are clear.
- The equalities at the ends are clear. Using the factorial formulae for computing binomial coefficients, we see that C ( n , k 1 ) = ( k / ( n k + 1 ) ) C ( n , k )
If we consider a sample k as k = 2 . . . k = 5 and we consider n = 7.
Then ( n 3 ) = ( n n / 2 ) because the floor value of n/2 is 3. But the same holds good on the right side of the equation with greater than signs with ( n 4 ) = ( n n / 2 ) because the ceiling value of n/2 is 4.
It is confusing to understand which of the steps is correct between the 1 s t and 2 n d provided in the option list. Also what is the first step for solving this, the problem definition or something similar does not exist in the text provided.
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Answers (2)

Caiden Barrett
Answered 2022-06-30 Author has 20 answers
Step 1
I got the answer wrong. Here is the answer :
Using the factorial formulae for computing binomial coefficients, we see that C ( n , k 1 ) = ( k / ( n k + 1 ) ) C ( n , k )
If k n / 2 , t h e n , k / ( n k + 1 ) < 1, so the "less than" signs are correct. Similarly, if k > n / 2 , t h e n , k / ( n k + 1 ) > 1, so the "greater than" signs are correct.
Step 2
Since C ( n , r ) = C ( n , n r ) , we have   C ( n , n 2 ) = C ( n , n n 2 ) = C ( n , n 2 )   because   n 2 + n 2 = n
Thus, the middle equality is clear.
The equalities at the ends are clear.
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Joshua Foley
Answered 2022-07-01 Author has 3 answers
Step 1
( n k + 1 ) ( n k ) = n ! ( n k 1 ) ! ( k + 1 ) ! n ! ( n k ) ! k ! = n ! ( n k 1 ) ! ( k 1 ) ! ( 1 k + 1 1 n k )
Step 2
Thus ( n k + 1 ) ( n k )  if and only if  k + 1 n k , i.e.  2 k + 1 n.
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