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2^1 + 2^2 + 2^3 + 2^4 + 2^5=

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Expert Post
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2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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New post 05 Nov 2017, 02:42
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A
B
C
D
E

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  15% (low)

Question Stats:

82% (01:02) correct 18% (00:58) wrong based on 95 sessions

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Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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New post 05 Nov 2017, 03:02
Bunuel wrote:
\(2^1 + 2^2 + 2^3 + 2^4 + 2^5=\)

(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2


\(2^1 + 2^2 + 2^3 + 2^4 + 2^5\) =\(2(1 + 2) + 2^3 + 2^4 + 2^5\) =\(2(3 + 2^2) + 2^4 + 2^5\) =\(2(7 + 2^3) + 2^5\)

=\(2(15) + 2^5\) =\(2(15 + 2^4)\) =\(2(31) = 62 = 2^6 - 2\) (Option E)
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New post 05 Nov 2017, 03:40
It's a GP series with r=2, a = 2.. Sum =a(r^n - 1)/r-1

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Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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New post 05 Nov 2017, 03:46
2+4+8+16+32=62

2^6-2=62

Ans:E

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2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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New post 05 Nov 2017, 04:02
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Bunuel wrote:
\(2^1 + 2^2 + 2^3 + 2^4 + 2^5=\)

(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2



HI...

a different way and a point to remember.....
\(2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^{n+1}-2\)

and this is WHY?
\(2^1+2^1=2^2.......2^1=2^2-2^1\)....

substitute this value in equation below

\((2^1 )+ 2^2 + 2^3 .........+ 2^{n-1} + 2^n=(2^2-2^1) + 2^2 + 2^3 .........+ 2^{n-1} + 2^n\)..

\(2^2-2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^2 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n-2^1\).....

\(2* 2^2 + 2^3 .........+ 2^{n-1} + 2^n-2^1= 2^3 + 2^3 .........+ 2^{n-1} + 2^n-2^1=2*2^3 .........+ 2^{n-1} + 2^n-2^1\)

and so on till \(2* 2^{n-1} + 2^n-2^1=2^n+2^n-2=2^{n+1}-2\)

so \(2^1+2^2+2^3+2^4+2^5=2^{5+1}-2=2^6-2\)

E

Bunuel, may be the choices could be put in ascending or descending order. choice A and B could be interchanged
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Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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New post 07 Nov 2017, 01:44
Bunuel wrote:
\(2^1 + 2^2 + 2^3 + 2^4 + 2^5=\)

(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2


\(2^1 + 2^2 + 2^3 + 2^4 + 2^5=
2(1+2+2^2+2^3+2^4)=
2(3+4+8+16)=
2(31)=62\)

\(16*4=2^4*2^2=2^6=64\)

Therefore, E is the correct answer \(2^6-2\)
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Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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New post 08 Nov 2017, 17:43
Bunuel wrote:
\(2^1 + 2^2 + 2^3 + 2^4 + 2^5=\)

(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2


We can evaluate the given equation:

2^1 + 2^2 + 2^3 + 2^4 + 2^5

2 + 4 + 8 + 16 + 32 = 62

We see that 62 = 64 - 2, which is 2^6 - 2.

Answer: E
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Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5=   [#permalink] 08 Nov 2017, 17:43
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