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# Sum!

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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
+878
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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
+23593
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I get    2^9 / 10!       (a,b) = (9,10)

Dec 28, 2018
#3
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deleted

Dec 28, 2018
edited by Guest  Dec 28, 2018
#4
+111328
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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)

Dec 28, 2018
edited by CPhill  Dec 28, 2018
#5
+878
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Thank you, guys!

Dec 28, 2018