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If (n+2)!= n!(an^2+bn+c), then abc=?

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by Max@Math Revolution » Sun Mar 11, 2018 11:57 pm
[GMAT math practice question]

If (n+2)!= n!(an^2+bn+c), then abc=?

A. 2
B. 3
C. 4
D. 6
E. 8
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by GMATGuruNY » Mon Mar 12, 2018 2:12 am
Max@Math Revolution wrote:[GMAT math practice question]

If (n+2)!= n!(an^2+bn+c), then abc=?

A. 2
B. 3
C. 4
D. 6
E. 8
(n+2)! = n!(an² + bn + c)

(n+2)!/n! = an² + bn + c

[(n+2)(n+1)(n!)]/n! = an² + bn + c

(n+2)(n+1) = an² + bn + c

n² + 3n + 2 = an² + bn + c.

Thus:
a=1, b=3, and c=2, with the result that abc = 1*3*2 = 6.

The correct answer is D.
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by Max@Math Revolution » Tue Mar 13, 2018 11:17 pm
=>

(n+2)! = (n+2)(n+1)n! = (n^2 + 3n + 2)n!
So,
a = 1, b = 3, and c = 2.
Thus, abc = 6.

Therefore, D is the answer.

Answer: D
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by Brent@GMATPrepNow » Wed Mar 14, 2018 8:06 am
Max@Math Revolution wrote:[GMAT math practice question]

If (n + 2)! = n!(an² + 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

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by deloitte247 » Sun Mar 18, 2018 9:45 am
$$\left(n+2\right)!=n!\left(an^2+bn+c\right)$$
$$\sin ce\ \left(n+2\right)!\ can\ be\ reduced\ to\ give\ \left(n+2\right)\left(n+1\right)n!$$
$$Then\ we\ have,\ \left(n+2\right)\left(n+1\right)n!=n!\left(an^2+bn+c\right)$$
$$Dividing\ through\ by\ n!,\ we\ have\ $$
$$\left(n+2\right)\left(n+1\right)=an^2+bn+c$$
$$n^2+3n+2=an^2+bn+c$$
$$Comparing\ the\ coefficients,\ we\ can\ see\ that$$
$$a=1\ \left(coefficient\ of\ n^2\right),\ b=3\ \left(coefficient\ of\ n\right)\ and\ c=2\ $$
$$Therefore,\ abc=1\cdot3\cdot2=6$$
$$Hence,\ the\ correct\ option\ is\ D$$
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