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n!(an^2+bn+c)

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Carcass
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n!(an^2+bn+c) [#permalink] New post   13 Oct 2021, 07:05
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If \((n+2)!= n!(an^2+bn+c),\) then \(abc=?\)

\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)
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D
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GreenlightTestPrep
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Re: n!(an^2+bn+c) [#permalink] New post   13 Oct 2021, 08:37
2
Carcass wrote:
If \((n+2)!= n!(an^2+bn+c),\) then \(abc=?\)

\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)

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
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Re: n!(an^2+bn+c) [#permalink] New post   21 Jun 2024, 09:01
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Re: n!(an^2+bn+c) [#permalink]
21 Jun 2024, 09:01
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