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# 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

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Manager
Joined: 13 Jul 2010
Posts: 139
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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Updated on: 05 Dec 2013, 01:57
3
24
00:00

Difficulty:

55% (hard)

Question Stats:

64% (01:58) correct 36% (02:24) wrong based on 462 sessions

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2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

A. 2^22(2^(23)-1)

B. 2^22(2^(24)-1)

C. 2^22(2^(25)-1)

D. 2^23(2^(21)-1)

E. 2^23(2^(22)-1)

Originally posted by gettinit on 21 Nov 2010, 20:34.
Last edited by Bunuel on 05 Dec 2013, 01:57, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Veritas Prep and Orion Instructor
Joined: 26 Jul 2010
Posts: 309

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22 Nov 2010, 12:18
19
19
Hey guys,

GREAT question - thanks for sharing this! This one is a terrific platform for what I call the Three Guiding Principles for Exponents:

1) Find common bases (which usually means you have to factor out bases into prime factors)

2) Multiply (which means that you have to factor out addition/subtraction terms to turn them into multiplication)

3) Find patterns (usually using small numbers to predict how patterns will look with the larger numbers you're given)

Here, I'll use Guiding Principles 2 and 3 (because all the bases are already 2):

All those terms - 2^22 through 2^44 - are going to be too much to carry around, so I'll use a smaller set to try to establish a pattern. Let's say it was just:

2^22 + 2^23 + 2^24 + 2^25

Then we'd need to use Guiding Principle #2 to turn this addition into multiplication, by factoring out the common 2^22 from all terms:

2^22 (1 + 2^1 + 2^2 + 2^3)

2^22 (1 + 2 + 4 + 8)

2^22 (15)

Looking at the answer choices we know we want to have 2^exponent - 1 in there, and we can make the 15 here 16-1, or 2^4 - 1, so we have:

2^22 (2^4 - 1) ----> Started with 4 terms (2^22 through 2^25) and ended up with the 2^22 multiplied by 2^4 - 1

Let's try again, but with five terms to start to see how the pattern goes:

2^22 + 2^23 + 2^24 + 2^25 + 2^26

Factor to:

2^22 (1 + 2 + 2^2 + 2^3 + 2^4)

2^22 (1 + 2 + 4 + 8 + 16)

2^22 (31)

And if we put it in that form as the answer choices we get:

2^22 (2^5 - 1) ----> starting with 5 terms, we end up with 2^22 multiplied by 2^5 - 1

Looking at the pattern, it should become fairly clear that we'll always end up with 2^22 (what we factor) multiplied by 2^an exponent equal to the number of terms we started with, minus 1. So the number of terms that we start with will equal the exponent in 2^x - 1.

Since we'll have 2^22 through 2^44, we have 44-22 + 1 terms (it's an inclusive set), so 23 terms total, and we'll end up with:

2^22 * (2^23 - 1)

_________________

Brian

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##### General Discussion
SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1855
Concentration: General Management, Nonprofit

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21 Nov 2010, 21:18
2
gettinit wrote:
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

$$2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)$$

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
$$a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)$$

Thus I think the final answer is B.
Manager
Joined: 20 Apr 2010
Posts: 217
WE 1: 4.6 years Exp IT prof

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22 Nov 2010, 12:06
Answer should be A not B
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SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1855
Concentration: General Management, Nonprofit

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22 Nov 2010, 12:32
Lol, sorry, that was my mistake. Answer is A. Thanks Brian!
Manager
Joined: 13 Jul 2010
Posts: 139

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22 Nov 2010, 20:28
VeritasPrepBrian wrote:
Hey guys,

GREAT question - thanks for sharing this! This one is a terrific platform for what I call the Three Guiding Principles for Exponents:

1) Find common bases (which usually means you have to factor out bases into prime factors)

2) Multiply (which means that you have to factor out addition/subtraction terms to turn them into multiplication)

3) Find patterns (usually using small numbers to predict how patterns will look with the larger numbers you're given)

Here, I'll use Guiding Principles 2 and 3 (because all the bases are already 2):

All those terms - 2^22 through 2^44 - are going to be too much to carry around, so I'll use a smaller set to try to establish a pattern. Let's say it was just:

2^22 + 2^23 + 2^24 + 2^25

Then we'd need to use Guiding Principle #2 to turn this addition into multiplication, by factoring out the common 2^22 from all terms:

2^22 (1 + 2^1 + 2^2 + 2^3)

2^22 (1 + 2 + 4 + 8)

2^22 (15)

Looking at the answer choices we know we want to have 2^exponent - 1 in there, and we can make the 15 here 16-1, or 2^4 - 1, so we have:

2^22 (2^4 - 1) ----> Started with 4 terms (2^22 through 2^25) and ended up with the 2^22 multiplied by 2^4 - 1

Let's try again, but with five terms to start to see how the pattern goes:

2^22 + 2^23 + 2^24 + 2^25 + 2^26

Factor to:

2^22 (1 + 2 + 2^2 + 2^3 + 2^4)

2^22 (1 + 2 + 4 + 8 + 16)

2^22 (31)

And if we put it in that form as the answer choices we get:

2^22 (2^5 - 1) ----> starting with 5 terms, we end up with 2^22 multiplied by 2^5 - 1

Looking at the pattern, it should become fairly clear that we'll always end up with 2^22 (what we factor) multiplied by 2^an exponent equal to the number of terms we started with, minus 1. So the number of terms that we start with will equal the exponent in 2^x - 1.

Since we'll have 2^22 through 2^44, we have 44-22 + 1 terms (it's an inclusive set), so 23 terms total, and we'll end up with:

2^22 * (2^23 - 1)

great explanation Brian, and for providing overall methodology on attacking these problems - immensely helpful.
Senior Manager
Joined: 08 Nov 2010
Posts: 341

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18 Feb 2011, 23:34
thanks, great explanation. +1
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SVP
Joined: 06 Sep 2013
Posts: 1745
Concentration: Finance

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04 Dec 2013, 17:53
whiplash2411 wrote:
gettinit wrote:
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

$$2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)$$

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
$$a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)$$

Thus I think the final answer is B.

Dude is that geometric progression formula correct?

Cheers
J
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

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04 Dec 2013, 21:04
jlgdr wrote:
whiplash2411 wrote:
gettinit wrote:
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

$$2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)$$

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
$$a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)$$

Thus I think the final answer is B.

Dude is that geometric progression formula correct?

Cheers
J

Sum of GP where first term is A, common ratio is R and total number of terms is n is given by
$$Sum = \frac{A(1 - R^n)}{(1 - R)}$$
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Karishma
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Intern
Joined: 28 Jan 2017
Posts: 4
Location: India
Concentration: Finance, General Management
WE: Securities Sales and Trading (Mutual Funds and Brokerage)
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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22 Mar 2017, 10:05
1
gettinit wrote:
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

A. 2^22(2^(23)-1)

B. 2^22(2^(24)-1)

C. 2^22(2^(25)-1)

D. 2^23(2^(21)-1)

E. 2^23(2^(22)-1)

A = 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) ... 1
2A = 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) + 2^45 ... 2

2 - 1 gives
A = 2^45 - 2^22 = 2^22 ( 2^23 - 1)
Manager
Status: GMAT...one last time for good!!
Joined: 10 Jul 2012
Posts: 60
Location: India
Concentration: General Management
GMAT 1: 660 Q47 V34
GPA: 3.5
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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22 Mar 2017, 10:21
dhavalpgk wrote:
gettinit wrote:
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

A. 2^22(2^(23)-1)

B. 2^22(2^(24)-1)

C. 2^22(2^(25)-1)

D. 2^23(2^(21)-1)

E. 2^23(2^(22)-1)

A = 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) ... 1
2A = 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) + 2^45 ... 2

2 - 1 gives
A = 2^45 - 2^22 = 2^22 ( 2^23 - 1)

Good job! Nice way of going about the problem!
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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) = &nbs [#permalink] 22 Mar 2017, 10:21
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