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2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=

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agalstia
 Post subject: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Thu Sep 30, 2010 12:48 am 
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Course Students


Posts: 12
Since all the bases are the same, I merely added up all the exponents to equal 37, but that is the wrong answer. I also tried to factor out a 2^2 from each number, but I was lost as to which direction to go afterwards.

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=


A) 2^9
B) 2^10
C) 2^16
D) 2^35
E) 2^37

OA is A. This is from the GMAT Prep software.
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rajivbhatia2007
 Post subject: Re: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Thu Sep 30, 2010 1:08 am 
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Students


Posts: 8
I guess there are different ways you could do this problem:
For GMAT, you probably should memorize the powers of 2 up to 2^10. Adding it out

2+2+4+8+16+32+64+128+256

The answer choices:

a) 2^9 = 512
b) 2^10 = 1024
c) 2^16 = (2^6)(2^10) = 64 x 1024 approx 64,000 (by looking at the addition, clearly this number is too big)
d) clearly too big
e) clearly too big

By adding up the choices, the answer is A


The other way to do this is:
2, 2, 4, 8, 16, 32, 64, 128, 256
Each term in the sequence is equal to the sum of the preceding terms. So the term 256 is the sum of 2 + 2 + 4 + .... + 128. Therefore: 2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 = 256 + 256 = 512 = 2^9

best,
rajiv
my gmat experience: http://gmathints.com/

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Rajiv Bhatia
www.gmathints.com
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agalstia
 Post subject: Re: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Thu Sep 30, 2010 1:29 am 
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Course Students


Posts: 12
thanks. that makes sense. the first approach just seems like it would take longer than 2 minutes.
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jssaggu.tico
 Post subject: Re: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Fri Oct 01, 2010 9:18 pm 
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Course Students


Posts: 15
The sum of the first n terms of a geometric progression is:
a(|r^n-1|)/|1 – r| ..................1
for the following type of series

Sn = a + ar + ar2 + … + ar^(n-1)=a(1+r+r^2+r^3 … +r^(n-1))

We have 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^7+2^8
=2+2(1+2+2^2+2^3+2^4+.....2^7)
=2+2(2^8-1)=2^9---by using 1

Just remember the series formula, and rest is cup of cake. It really helps when we have much complex geometric series or in case, we have general series with terms of 'n'
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tim
 Post subject: Re: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Fri Oct 08, 2010 1:41 am 
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ManhattanGMAT Staff


Posts: 2166
Location: Southwest Airlines, seat 21C
thanks, Navjot..

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Manhattan GMAT Instructor
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sarahmailings
 Post subject: Re: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Tue Nov 02, 2010 10:44 am 
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Course Students


Posts: 7
navjotsingh05 wrote:
The sum of the first n terms of a geometric progression is:
a(|r^n-1|)/|1 – r| ..................1
for the following type of series

Sn = a + ar + ar2 + … + ar^(n-1)=a(1+r+r^2+r^3 … +r^(n-1))

We have 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^7+2^8
=2+2(1+2+2^2+2^3+2^4+.....2^7)
=2+2(2^8-1)=2^9---by using 1

Just remember the series formula, and rest is cup of cake. It really helps when we have much complex geometric series or in case, we have general series with terms of 'n'


I'm not quite sure I understand the series formula explanation. I did find some of the explanations posted on this forum quite helpful. Maybe others will too.
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graphica
 Post subject: Re: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Wed Nov 03, 2010 11:33 am 
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Students


Posts: 5
It is a bit lengthy but can done.

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

Rewrite the first 2+2 as 2^2

Therefore

2^2[1+1+2+2^2+2^3+2^4+2^5+2^6] Now making 1+1+2=2^2
2^2[2^2{1+1+2+2^2+2^3+2^4}] Now making 1+1+2=2^2
2^2[2^2{2^2(1+1+2+2^2)}] Now making 1+1+2=2^2
2^2[2^2{2^2(2^2+2^2)}] Now making 1+1+2=2^2
2^2[2^2{2^2(2^2(1+1))}] Now making 1+1=2
Finally it comes to
2^2 X 2^2 X 2^2 X 2^2 X 2

Adding the powers since the base is the same makes it
2^2+2+2+2+1 = 2^9

I hope it helps.
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mschwrtz
 Post subject: Re: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
 Post Posted: Fri Nov 05, 2010 2:08 am 
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ManhattanGMAT Staff


Posts: 506
Alternatively you could look for a pattern.

2+2=2^2

2+2+2^2=2^3

so you can see that each term in the expression is equal to the sum of all the previous terms, so that adding each term is doubling the previous sum. Doubling=multiplying by 2=increasing the power by 1.
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