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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
((n-1)!+n!+(n+1)!)/n!
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15 Nov 2021, 06:27
15 Nov 2021, 06:27
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\(\frac{(n-1)!+n!+(n+1)!}{n!}\)
Which of the following values of n will result between 16 and 17 for the equation?
A. 14
B. 15
C. 16
D. 17
E. 18
Which of the following values of n will result between 16 and 17 for the equation?
A. 14
B. 15
C. 16
D. 17
E. 18
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
((n-1)!+n!+(n+1)!)/n!
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15 Nov 2021, 06:58
15 Nov 2021, 06:58
GeminiHeat wrote:
\(\frac{(n-1)!+n!+(n+1)!}{n!}\)
Which of the following values of n will result between 16 and 17 for the equation?
A. 14
B. 15
C. 16
D. 17
E. 18
Which of the following values of n will result between 16 and 17 for the equation?
A. 14
B. 15
C. 16
D. 17
E. 18
Useful fraction property: \(\frac{a+b+c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\)
Apply the property to get: \(\frac{(n-1)!+n!+(n+1)!}{n!}=\frac{(n-1)!}{n!}+\frac{n!}{n!}+\frac{(n+1)!}{n!}\)
\(=\frac{1}{n}+1+(n+1)\)
\(=n + 2 + \frac{1}{n}\)
From the answer choices, we can see that the value of n must be an integer from 14 to 18 inclusive, which means the fraction \(\frac{1}{n}\) must be between 0 and 1.
This means the expression \(n + 2 + \frac{1}{n}\) will be between 16 and 17 when \(n = 14\)
Answer: A
((n-1)!+n!+(n+1)!)/n!
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17 Nov 2021, 01:35
17 Nov 2021, 01:35
1
\(\frac{(n-1)!+n!+(n+1)!}{n!} = \frac{(n-1)!}{n!} + \frac{n!}{n!} + \frac{(n+1)!}{n!} = \frac{(n-1)!}{n*(n-1)!} + 1 + \frac{(n+1)*n!}{n!} = \frac{1}{n} + 1 + (n+1) = n + 2 + \frac{1}{n}\)
Since all options are greater than 1, \(0< \frac{1}{n} < 1\)
We only need to check the term \((n+2)\) to find the answer.
Except for 14 all other options go above \(17\).
Hence, Answer is A
Since all options are greater than 1, \(0< \frac{1}{n} < 1\)
We only need to check the term \((n+2)\) to find the answer.
Except for 14 all other options go above \(17\).
Hence, Answer is A
