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3^n+3^{n+1}+3^{n+2}/3^{n-2}+3^{n-1}+3^n=?

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3^n+3^{n+1}+3^{n+2}/3^{n-2}+3^{n-1}+3^n=?

by Max@Math Revolution » Tue Jun 26, 2018 1:21 am

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[GMAT math practice question]


$$\frac{3^n+3^{n+1}+3^{n+2}}{3^{n-2}+3^{n-1}+3^n}\ =\ ?$$

A. 3^-2
B. 3^-n
C. 3^n
D. 3^2
E. 1
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by GMATGuruNY » Tue Jun 26, 2018 4:39 am
Max@Math Revolution wrote:[GMAT math practice question]


$$\frac{3^n+3^{n+1}+3^{n+2}}{3^{n-2}+3^{n-1}+3^n}\ =\ ?$$

A. 3^-2
B. 3^-n
C. 3^n
D. 3^2
E. 1
Let n = 3.
Plugging n = 3 into the given expression, we get:
$$\frac{3^3+3^{3+1}+3^{3+2}}{3^{3-2}+3^{3-1}+3^3}\\$$

$$\frac{3^3+3^4+3^5}{3^1+3^2+3^3}\\$$

$$\frac{3^3(1 + 3 + 3^2)}{3(1+3+3^2)} = 3^2\\$$

When n=3 is plugged into the answers, the result must be 3².
Only D is viable.

The correct answer is D.
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by Max@Math Revolution » Thu Jun 28, 2018 2:20 am
=>

$$\frac{\left(3^n+3^{n+1}+3^{n+2}\right)}{\left(3^{n-2}+3^{n-1}+3^n\right)}\ =\ \frac{\left(3^n\left(1+3^1+3^2\right)\right)}{\left(3^{n-2}\left(1+3^1+3^2\right)\right)}=\frac{\left(3^{2+\left(n-2\right)}\cdot13\right)}{\left(3^{n-2}\cdot13\right)}=3^2$$

We can also find the answer by plugging in n = 2. This gives (3^2 + 3^3 + 3^4)/(3^0 + 3^1 + 3^2) = (9 + 27 + 81)/(1 + 3 + 9) = 107/13 = 9 = 3^2.

In addition, we can plug-in number 2.
Then we have (32 + 33 + 34)/(30 + 31 + 32) = (9+27+81)/(1+3+9) = 107/13 = 9.

Therefore, the answer is D.

Answer: D
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by GMATGuruNY » Thu Jun 28, 2018 2:29 am
We can also find the answer by plugging in n = 2. This gives (3^2 + 3^3 + 3^4)/(3^0 + 3^1 + 3^2) = (9 + 27 + 81)/(1 + 3 + 9) = 107/13 = 9 = 3^2.
Both C and D yield a value of 3² when n=2.
As a result, a test-taker would have to test another case to determine whether the correct answer is C or D.
When plugging in values, we should avoid numbers that appear in the answer choices.
Here -- since 2 appears in the answer choices -- we should test a value other than 2.
When n=3, the only viable answer choice is D, as shown in my earlier solution.
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by Jeff@TargetTestPrep » Thu Jun 28, 2018 6:28 am
Max@Math Revolution wrote:[GMAT math practice question]


$$\frac{3^n+3^{n+1}+3^{n+2}}{3^{n-2}+3^{n-1}+3^n}\ =\ ?$$

A. 3^-2
B. 3^-n
C. 3^n
D. 3^2
E. 1
The common factor in the numerator is 3^n, and the common factor in the denominator is 3^(n - 2). Thus, we can factor the numerator and denominator as follows:

[3^n(1 + 3 + 3^2)] / [3^(n - 2)(1 + 3 + 3^2)]

3^n / 3^(n - 2)

3^(n - (n - 2))

3^2

Answer: D

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