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avatar+853 

The sum \(\frac{1}{1! 9!} + \frac{1}{3! 7!} + \frac{1}{5! 5!} + \frac{1}{7! 3!} + \frac{1}{9! 1!}\)
can be expressed in the form \(\frac{2^a}{b!},\) where \(a\) and \(b\) are positive integers. Enter the ordered pair \((a,b).\)

 Dec 28, 2018
 #1
avatar+853 
+1

I think I got it is it \(\frac{2^1}{14175}\), but how would you express \(b?\)

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 Dec 28, 2018
 #2
avatar+17163 
+2

I get    2^9 / 10!       (a,b) = (9,10)

 Dec 28, 2018
 #3
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+1

deleted

 Dec 28, 2018
edited by Guest  Dec 28, 2018
 #4
avatar+98005 
+2

Note that  we can write

 

10!  [ 1 / (9! * 1! )  + 1 / (3!*7! )  + 1/ (5!*5!) + 1/ (7!*3!) + 1/(9!*1!) ]  =

 

10! / (9! * 1!) + 10! / (3!*7!) + 10! / (5!*5!) + 10! / (7!*3!) + 10!/(9!*1!)  

 

C(10, 1)  + C(10, 3) + C(10, 5)  + C(10,3) + C(10,1)  =

 

2 C(10, 1)  + 2 (10, 3) + C(10, 5)   =

 

20  + 240   +  252  =

 

512

 

 

So

 

10!  [ 1 / (9! * 1! )  + 1 / (3!*7!  + 1/ (5!*5!) + 1/ (7!*3!) + 1/(9!*1!) ]  =  512

 

[ 1 / (9! * 1! )  + 1 / (3!*7!  + 1/ (5!*5!) + 1/ (7!*3!) + 1/(9!*1!) ]  =  512 / 10!  =  29 / 10!

 

(a , b)  =  (9, 10)

 

 

cool cool cool

 Dec 28, 2018
edited by CPhill  Dec 28, 2018
 #5
avatar+853 
+2

Thank you, guys!

 Dec 28, 2018

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