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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=?  [#permalink]

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New post 26 Jun 2018, 02:30
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

83% (01:22) correct 18% (01:23) wrong based on 52 sessions

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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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Re: 3^n+3^{n+1}+3^{n+2}/3^{n-2}+3^{n-1}+3^n=?  [#permalink]

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New post 26 Jun 2018, 02:44
I just assumed n=1 and solved

In My Opinion
Ans: D
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3^n+3^{n+1}+3^{n+2}/3^{n-2}+3^{n-1}+3^n=?  [#permalink]

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New post 26 Jun 2018, 02:56
MathRevolution wrote:
[GMAT math practice question]

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

\(A. 3^{-2}\)
\(B. 3^{-n}\)
\(C. 3^n\)
\(D. 3^2\)
\(E. 1\)


two ways..

1) proper method

\(\frac{3^n+3^{n+1}+3^{n+2}}{3^{n-2}+3^{n-1}+3^n}=\frac{3^n(1+3+3^2)}{3^{n-2}(1+3+3^2)}=\frac{3^n}{3^{n-2}}=3^{n-n+2}=3^2\)
D

2) put n as >=2, so that DENOMINATOR does not have any fraction, leading to some extra calculations
..
\(\frac{3^n+3^{n+1}+3^{n+2}}{3^{n-2}+3^{n-1}+3^n}=\frac{3^2+3^{2+1}+3^{2+2}}{3^{2-2}+3^{2-1}+3^2}=\frac{3^2(1+3+3^2)}{1+3+3^2}=3^2\)
Here both C and D give us 3^2, so we can try with n=3...


D
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Re: 3^n+3^{n+1}+3^{n+2}/3^{n-2}+3^{n-1}+3^n=?  [#permalink]

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New post 26 Jun 2018, 05:56
Quote:
MathRevolution wrote:
[GMAT math practice question]

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

\(A. 3^{-2}\)
\(B. 3^{-n}\)
\(C. 3^n\)
\(D. 3^2\)
\(E. 1\)


2) put n as 2, so that DENOMINATOR does not have any fraction, leading to some extra calculations
..
\(\frac{3^n+3^{n+1}+3^{n+2}}{3^{n-2}+3^{n-1}+3^n}=\frac{3^2+3^{2+1}+3^{2+2}}{3^{2-2}+3^{2-1}+3^2}=\frac{3^2(1+3+3^2)}{1+3+3^2}=3^2\)

D


Both C and D yield \(3^2\) 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.
Plug n=3 into the given expression:

\(\frac{3^3+3^{3+1}+3^{3+2}}{3^{3-2}+3^{3-1}+3^3}\)

\(\frac{3^3+3^4+3^5}{3^1+3^2+3^3}\)

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

\(3^2\)

The correct answer must yield \(3^2\) when n=3.
In this case, only D is viable.


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Re: 3^n+3^{n+1}+3^{n+2}/3^{n-2}+3^{n-1}+3^n=?  [#permalink]

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New post 28 Jun 2018, 03:35
=>

\(\frac{(3^n + 3^{n+1} + 3^{n+2})}{(3^{n-2} + 3^{n-1} + 3^n)}\)
\(= \frac{(3^n(1 + 3^1 + 3^2))}{(3^{n-2}(1 + 3^1 + 3^2 ))}\)
\(= \frac{(3^{2+(n-2)}*13)}{(3^{n-2}*13)}\)
\(= 3^2\)

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

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

Therefore, the answer is D.

Answer: D
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Re: 3^n+3^{n+1}+3^{n+2}/3^{n-2}+3^{n-1}+3^n=? &nbs [#permalink] 28 Jun 2018, 03:35
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