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

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

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New post 06 Nov 2017, 01:39
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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 \(2^{23}(2^{(22)}-1)\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Nov 2017, 01:46, edited 1 time in total.
Edited the question and added the OA.

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

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New post 06 Nov 2017, 02:02
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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, \(2^{22}(1+2+2^2+...+2^{22})=2^{22}(2^{23} - 1)\).

Answer: A.

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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New post 06 Nov 2017, 02:15
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 \(2^{23}(2^{(22)}-1)\)


Its a GP Series, a = 2^22, r = 2, n=23

S = a(r^n-1)/r-1

Kudos [?]: 5 [0], given: 120

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

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New post 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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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New post 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)

Answer: A
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Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =   [#permalink] 08 Nov 2017, 17:26
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