# NMAT Practice Exam Math Problem: $3^{n+2}+(3^{n+3}-3^{n+1}) =~?$

The problem is from an NMAT Practice Exam. The problem is multiple choice. It looks easy enough...

$$3^{n+2}+(3^{n+3}-3^{n+1}) =~?$$

a.) $\dfrac1{3^{n+1}}$

b.) $\dfrac1{3^{n+2}}$

c.) $\dfrac38$

d.) $\dfrac13$

The answer given is $\frac13$, but I don't know how they got that.

My attempts:

$$3^{n+2}+(3^{n+3}-3^{n+1})=3^n(9+27-3)=33\cdot3^n$$

Another attempt using self similarity...

$$y=3^{n+2}+(3^{n+3}-3^{n+1})$$

$$3y=3^{n+3}+(3^{n+4}-3^{n+2})$$

$$3y-y=3^{n+4}-2\cdot3^{n+2}+3^{n+1}$$

I'm trying help someone out with the math section, but I'm lost on how to solve this one. Thanks in advance.

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The answer I get is not one of your options. Have I misinterpreted the question? – Austin Mohr Nov 30 '12 at 7:22
I just looked up and browsed through an NMAT practice exam, and every question on it had 5 options. Are you sure that this one only has 4? – rayradjr Nov 30 '12 at 7:49
thanks for replying. I think Henry figured it out. I think the + sign is supposed to be a divide symbol. Solving it that way gives an answer of 3/8. Sorry, for the trouble. – A.G. Nov 30 '12 at 7:52

If you are allowed to assume a typo of $+$ for the similar $\div$, you can get one of the choices:
$$3^{n+2}\div(3^{n+3}-3^{n+1}) = \frac38.$$
That was my notion, too; unfortunately, it’s the wrong choice, and I can’t see any reasonable typo or OCR error that would make $\frac13$ the right choice. I agree that the answer choices pretty much guarantee that the original problem had a division or negative exponent somewhere. – Brian M. Scott Nov 30 '12 at 7:42
@Austin Mohr: But $3^{n+2}\div(3^{n+3}-3^{n+1})=\dfrac{3^{n+1}\times 3}{3^{n+1}\times (3^2-1)}$ – Henry Nov 30 '12 at 8:00