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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
The sequence f(n) = (2n)! ÷ n! is defined for all positive integer
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17 Aug 2021, 09:11
17 Aug 2021, 09:11
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Question Stats:
47% (03:14) correct
52% (03:37) wrong based on 17 sessions
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The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n. If x is defined as the product of the first 10 ten terms of this sequence, which of the following is the greatest factor of x?
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
The sequence f(n) = (2n)! ÷ n! is defined for all positive integer
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06 Sep 2021, 08:18
06 Sep 2021, 08:18
2
GeminiHeat wrote:
The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n. If x is defined as the product of the first 10 ten terms of this sequence, which of the following is the greatest factor of x?
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55
Let's look for a pattern...
Term 1 = 2!/1! = 2 = 2^1
Term 2 = 4!/2! = 12 = 3(2^2)
Term 3 = 6!/3! = 120 = 15(2^3)
Term 4 = 8!/4! = 1680 = 105(2^4)
Continuing we get...
Term 9 = 18!/9! = ???(2^9)
Term 10 = 20!/1! = ???(2^10)
So, if we focus on the 2's, we get: x = (2^1)(2^2)(2^3)(2^4). . . (2^9)(2^10)
= 2^(1+2+3+4+5+6+7+8+9+10)
= 2^(55)
Answer: E
General Discussion
Re: The sequence f(n) = (2n)! ÷ n! is defined for all positive integer
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20 Aug 2021, 15:39
20 Aug 2021, 15:39
1
(2n)!/n! = (1*2*3*......*2n)/n! = (1*3*5*7*......*2n-1) * (2*4*6*8......*2n)/n! = (1*3*5*......*2n-1) * (1*2*3......*n)*2^n /n!
For n=1, 2^n has a power of 1. Similarly, for n=10, 2^n will have a power of 10. We have to add all the powers from n=1 to n=10
Answer= 2^(1+2+3+4+.........+10)=2^55
For n=1, 2^n has a power of 1. Similarly, for n=10, 2^n will have a power of 10. We have to add all the powers from n=1 to n=10
Answer= 2^(1+2+3+4+.........+10)=2^55
