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Surds and Indices Practice Test 2

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Question : 7 of 17

Marks: +1, -0

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

is equivalent to:

32
39
35
38

Solution:

Concept:

Use exponent laws

a^{m+n}=a^m a^n

and

a^{-m}=\frac{1}{a^m}

to simplify the given expression.

Explanation:

Given:

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

.

Factor the numerator using

3^n

:

3^{n+3}-3^{n+1}+3^n=3^n(3^3-3+1)

.

Simplify the bracket:

3^3-3+1=27-3+1=25

, so numerator

=25\cdot 3^n

.

Write the denominator:

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

.

Cancel

3^n

:

\frac{25\cdot 3^n}{\frac{2}{3}\cdot 3^n}=\frac{25}{2/3}=\frac{75}{2}

.

Now add

2^{-1}=\frac{1}{2}

:

\frac{75}{2}+\frac{1}{2}=\frac{76}{2}=38

.

Answer:

The value is

38

, so the correct option is D.

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