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

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

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Updated on: 08 Mar 2019, 08:07
4
32
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Difficulty:

55% (hard)

Question Stats:

64% (01:56) correct 36% (02:23) wrong based on 372 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, 21:34.
Last edited by Bunuel on 08 Mar 2019, 08:07, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Joined: 02 Sep 2009
Posts: 58441
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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06 Nov 2017, 02:02
3
3
arabella 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 $$2^{23}(2^{(22)}-1)$$

You can solve this question by applying the GP formula.

$$2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} =2^{22}(1+2+2^2+...+2^{22})=$$

Now, again you can apply the formula for GP or just observe the pattern:

$$1 + 2 = 3 =4-1 =2^2 - 1$$;
$$1 + 2 + 2^2 = 7 = 8-1=2^3 - 1$$;
$$1 + 2 + 2^2 + 2^3 = 15 = 16-1=2^4 - 1$$;
...
$$1+2+2^2+...+2^{22} = 2^{23} - 1$$.

Thus, $$2^{22}(1+2+2^2+...+2^{22})=2^{22}(2^{23} - 1)$$.

Similar question: https://gmatclub.com/forum/36-126078.html
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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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22 Nov 2010, 13:18
23
20
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)

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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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21 Nov 2010, 22:18
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.
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Concentration: Finance
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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04 Dec 2013, 18:53
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
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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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04 Dec 2013, 22:04
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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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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22 Mar 2017, 11:05
3
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)
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Posts: 7988
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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06 Nov 2017, 02:52
arabella 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 different way and a point to remember.....
$$2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^{n+1}-2$$

so...
$$2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} = (2^1+2^2+......2^{44})-(2^1+2^2+....+2^{21}$$
as per above formula
$$(2^1+2^2+......2^{44})-(2^1+2^2+....+2^{21}=(2^{(44+1)}-2)-(2^{(21+1)}-2)=2^{(44+1)}-2-2^{(21+1)}+2=2^{(45)}-2^{(22)}=2^{22}(2^{23}-1)$$
A

you can check why $$2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^{n+1}-2$$ at
https://gmatclub.com/forum/2-252919.html
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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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08 Nov 2017, 17:26
arabella 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 $$2^{23}(2^{(22)}-1)$$

Let’s begin by factoring out 2^22, which is common to each of the terms:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22)

To find an expression for 1 + 2 + 2^2 + … + 2^22, we will use the formula

a^n - 1 = (a - 1)(a^(n - 1) + a^(n - 2) + … + a + 1).

If we let a = 2 and n = 23, we obtain

2^23 - 1 = (2 - 1)(2^22 + 2^21 + … + 2 + 1).

Thus:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22) = 2^22(2^23 - 1)

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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

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23 Dec 2018, 09:35
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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =   [#permalink] 23 Dec 2018, 09:35
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