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avatar+14 

For the positive integer n, the rational number r satisfies \( \binom{2n}{n} - \binom{2n}{n + 1} = r \cdot \binom{2n}{n}.\)  Express r in terms of n (your problem should be as simplified as possible).

 Dec 18, 2023
 #1
avatar+222 
+1

We can simplify the equation:

\(\begin{align*} \binom{2n}{n} - \binom{2n}{n + 1} &= r * \binom{2n}{n} \ \Rightarrow \qquad \frac{(2n)!}{n!(n+1)!} - \frac{(2n)!}{(n+1)!(n)!} \\ &= r \cdot \frac{(2n)!}{n!(n!)} \ \Rightarrow \qquad \frac{2n-1}{(n+1)!} \\ &= r. \end{align*}\)

 

Therefore, the rational number r can be expressed in terms of n as:

r = (2n-1) / (n+1)!

 

This answer gives a simple and concise expression for r in terms of n, using factorial notation.

 Dec 18, 2023
 #2
avatar+14 
+1

Sorry, that's wrong.

 Dec 18, 2023
 #3
avatar+126665 
+1

   (2n)!               (2n)!                                                     (2n)!

________  -    ___________________     =       r  *   _________        rearrange as

(2n -n)! n!        (2n - (n + 1))!  (n + 1)!                          (2n - n)! n!

 

(2n)! (1 -r)                       (2n)!

_________  =    __________________            (divide out (2n)!   and    multiply through by -1 )        

(2n - n)! n!         (2n - (n + 1))! (n + 1)!

 

(1 - r)                               1

_________ =         _____________  

(n!) (n!)                   (n - 1)! (n + 1)!

 

( r - 1)                       -1

______  =     ________________

(n!)(n!)                (n -1)! ( n + 1)!

 

(r -1)  =     - (n! (n!)

             _____________

                (n - 1)! (n + 1)!

 

r - 1  =     -  (n)! / ( n -1)!   *   n! / (n + 1)!

 

r  -1  =    -     n  *     1/(n + 1)

 

r - 1  =   - n / (n + 1)

 

r  = 1 -  [  n / (n + 1)   ] 

 

r = ( n + 1  - n) / ( n + 1)

 

r =  1 / ( n + 1)

 

 

cool cool cool

 Dec 19, 2023
edited by CPhill  Dec 19, 2023

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