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

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Math Expert
Joined: 02 Sep 2009
Posts: 43867
2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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05 Nov 2017, 01:42
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$$2^1 + 2^2 + 2^3 + 2^4 + 2^5=$$

(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2
[Reveal] Spoiler: OA

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

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05 Nov 2017, 02:02
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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Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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05 Nov 2017, 02:40
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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05 Nov 2017, 02:46
2+4+8+16+32=62

2^6-2=62

Ans:E
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Joined: 02 Aug 2009
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2^1 + 2^2 + 2^3 + 2^4 + 2^5= [#permalink]

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05 Nov 2017, 03:02
Expert's post
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This post was
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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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07 Nov 2017, 00:44
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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08 Nov 2017, 16:43
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.

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Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5=   [#permalink] 08 Nov 2017, 16:43
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