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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]

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07 Sep 2013, 00:46
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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, 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]

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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) $$3*2^{n-1}$$

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]

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08 Sep 2013, 00:27
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]

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08 Sep 2013, 05:19
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Expert's post
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 $$2^{n-1}*2=2^n$$).

Similar questions:
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if-5-x-5-x-3-124-5-y-what-is-y-in-terms-of-x-109080.html

Hope this helps.
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink]

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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]

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27 Nov 2015, 09:03
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]

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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

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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ [#permalink]

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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^   [#permalink] 25 Oct 2017, 06:34
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