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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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    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.2012-11-30

1 Answers 1

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If you are allowed to assume a typo of $+$ for the similar $\div$, you can get one of the choices:

$3^{n+2}\div(3^{n+3}-3^{n+1}) = \frac38.$

Since all the choices have denominators, you need to do a division or use negative exponents to get them.

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    @BrianM.Scott Thanks for the reply. The answers for the practice exam are listed on a separate page. Only the letter for the correct answer is given. I think D was typed instead C by accident. sorry for the trouble.2012-11-30