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

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fozzzy
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If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post   Updated on: 07 Sep 2013, 06:31
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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) \(3*2^{n-1}\)
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E

Originally posted by fozzzy on 07 Sep 2013, 00:46.
Last edited by fozzzy on 07 Sep 2013, 06:31, 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, 03:18
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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   08 Sep 2013, 00:27
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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, 05:19
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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, 05:32
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] New post   27 Nov 2015, 09:03
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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) \(3*2^{n-1}\)


Generally the first thing to do in questions involving exponent is to take out common terms.

\(2^{n-1} + 2^n =\) = \(2^{n-1}*(1 + 2)\) = \(2^{n-1}*3\)
Option E
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post   28 Nov 2015, 12:57
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) \(3*2^{n-1}\)


Lets plug in some values of n

given
Quote:
n is a positive integer greater than 1


n = { 2,3,4,5.......}

If n = 2 then , \(2^{2-1} + 2^2\) = 6

If n = 3 then , \(2^{3-1} + 2^3\) = 12

From n =2 & 3 ; check the result is always a multiple of 3 as well as 2 , now check the options.

Among the given options only option (E) has 3 & 2

If n =2 then , \(3*2^{n-1}\) = 6

If n = 3 then , \(3*2^{n-1}\) = 12

So, answer is definitely (E)
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink] New post   21 Mar 2023, 02:48
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink]
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