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n!(an^2+bn+c)
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13 Oct 2021, 07:05
13 Oct 2021, 07:05
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Question Stats:
68% (01:12) correct
31% (01:01) wrong based on 16 sessions
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If \((n+2)!= n!(an^2+bn+c),\) then \(abc=?\)
\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)
\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: n!(an^2+bn+c)
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13 Oct 2021, 08:37
13 Oct 2021, 08:37
2
Carcass wrote:
If \((n+2)!= n!(an^2+bn+c),\) then \(abc=?\)
\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)
\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)
First recognize that (n + 2)! = (n + 2)(n + 1)(n)(n - 1)(n - 2)(n - 3)....(3)(2)(1)
= (n + 2)(n + 1)(n!)
= (n!)(n + 2)(n + 1)
The question tells us that (n + 2)! = n!(an² + bn + c)
So, we can write: (n!)(n + 2)(n + 1) = n!(an² + bn + c)
Divide both sides by n! to get: (n + 2)(n + 1) = an² + bn + c
Use FOIL to expand left side: n² + 3n + 2 = an² + bn + c
In other words: 1n² + 3n + 2 = an² + bn + c
So, a = 1, b = 3 and c = 2
This means abc = (1)(3)(2) = 6
Answer: D
Cheers,
Brent
Re: n!(an^2+bn+c)
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21 Jun 2024, 09:01
21 Jun 2024, 09:01
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
