+885 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 \((a,b).\)
+885 I think I got it is it \(\frac{2^1}{14175}\), but how would you express \(b?\)
+129899 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)
