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

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Director
Joined: 29 Nov 2012
Posts: 765
If n is a positive integer greater than 1, then 2^{n-1} + 2^  [#permalink]

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Updated on: 07 Sep 2013, 05:31
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Question Stats:

83% (01:10) correct 17% (01:14) wrong based on 165 sessions

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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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Originally posted by fozzzy on 06 Sep 2013, 23:46.
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]

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07 Sep 2013, 02:18
1
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]

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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]

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08 Sep 2013, 04:19
2
1
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]

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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]

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27 Nov 2015, 08:03
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, 11:57
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]

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25 Oct 2017, 05:34
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Re: If n is a positive integer greater than 1, then 2^{n-1} + 2^ &nbs [#permalink] 25 Oct 2017, 05:34
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# If n is a positive integer greater than 1, then 2^{n-1} + 2^

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