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.
$\endgroup$ 31 Answer
$\begingroup$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.
$\endgroup$ 4
