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If n is a positive integer greater than 1, then 2^{n-1} + 2^

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If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post 06 Sep 2013, 23:46
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If n is a positive integer greater than 1, then 2^{n-1} + 2^n =

A) 3^n
B) 2^n+1
C) 2^{2n-1}
D) 2^{n(n-1)}
E) 3*2^{n-1}
[Reveal] Spoiler: OA

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Last edited by fozzzy on 07 Sep 2013, 05:31, edited 1 time in total.
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Re: if n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post 07 Sep 2013, 02:18
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fozzzy wrote:
if n is a positive integer greater than 1, then 2^{n-1} + 2^n =
A) 3^n
B) 2^n+1
C) 2^{2n-1}
D) 2^{n(n-1)}
E) 3*2^{n-1}

Answer E

2^n-1+2^n
=>2^n-1(1+2)
=>2^n-1(3)
=>3*2^n-1
Option E
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post 07 Sep 2013, 23:27
I didn't understand the manipulation. Could you elaborate...
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post 08 Sep 2013, 04:19
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fozzzy wrote:
I didn't understand the manipulation. Could you elaborate...


If n is a positive integer greater than 1, then 2^{n-1} + 2^n =

A) 3^n
B) 2^n+1
C) 2^{2n-1}
D) 2^{n(n-1)}
E) 3*2^{n-1}

Factor out 2^{n-1}:

2^{n-1}(1+2)=3*2^{n-1} (notice that 2^{n-1}*2=2^n).

Answer E.

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Hope this helps.
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post 08 Sep 2013, 04:32
Here is what I did

2^n . 2^{-1}+ 2^n

we then get 2^n [ \frac{1}{2} + 1]

2^n [ \frac{3}{2}] --------> I stopped over here


so basically this step is valid-----> 2^n . 2^{-1} [3]

Add the exponents we get 2^{n-1}[3]
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^   [#permalink] 08 Sep 2013, 04:32
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