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algebra senior

Problem

\frac{2 ^{n + 4} - 2 ( 2 ^{n} )}{2 ( 2 ^{n + 3} )}

when simplified is:

(A)

2^{(} n + 1) - 1/8

(B)

- 2^{(} n + 1)

(C)

1 - 2^{n}

(D)

7/8

Solution

We have

2 (2^{n}) = 2^{n + 1}

, and

2 (2^{n + 3}) = 2^{n + 4}

. Thus,

\frac{2 ^{n + 4} - 2 ( 2 ^{n} )}{2 ( 2 ^{n + 3} )} = \frac{2 ^{n + 4} - 2 ^{n + 1}}{2 ^{n + 4}}

. Factoring out a

2^{n + 1}

in the numerator, we get

\frac{2 ^{n + 1} ( 2 ^{3} - 1 )}{2 ^{n + 4}} = \frac{8 - 1}{2 ^{3}} = \frac{7}{8}

.

Final answer

D