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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? 1) 2^9

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Manager
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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? 1) 2^9 [#permalink]

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29 May 2006, 16:46
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

1) 2^9
2) 2^10
3) 2^16
4) 2^35
5) 2^37
Manager
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29 May 2006, 16:53
tl372 wrote:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

1) 2^9
2) 2^10
3) 2^16
4) 2^35
5) 2^37

I think this is one of those tricky question...
I would first factor out 2^8 from the whole equation... I will then get

2^8*(2^-7+2^-7+2^-6+2^-5+2^-4+2^-3+(1/4)+(1/2)+1)

the value in read a small enought that they can be considered or assumed to be zeros. So 2^9 would the be logical answer.

Senior Manager
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29 May 2006, 17:39
Its like the sum of an infinite Geometric progression with an r of 1/2.
The answer is 2^9 if the series is 2^8+2^7+2^6+...+2^2+2^1+2^0

Since the given series is 2^2+2^1+2^0+1, the sum of the series is 2^9+1.

However since that answer is not provided, I'd go with A.
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29 May 2006, 21:29
tl372 wrote:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

1) 2^9
2) 2^10
3) 2^16
4) 2^35
5) 2^37

=2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
=2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
=2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
=2^4 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
=2^5 + 2^5 + 2^6 + 2^7 + 2^8
.
.
.
.
=2^8 + 2^8 = 2^9
Manager
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29 May 2006, 21:39
I also got A)

=2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
=4+4+8+16+32+64+128+256=512=256*2=2^9

It shouldnâ€™t take you more than 30 sec to sum (4+8+16...+256)

You can check the 2^9 by doing prime factorization of 512, but I think this problem is more of logic than of doing operations.

Cheers!!
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29 May 2006, 22:54
A for me too.......

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
=> 2(1+1+2+ 2^3 + 2^4 + 2^5 + 2^6 + 2^7)
=>2(2+2+ 2^3 + 2^4 + 2^5 + 2^6 + 2^7)

if you see the pattern here, you will come to the answer
Manager
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30 May 2006, 05:34
A for me too.. same approach as prof's..
Manager
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30 May 2006, 05:49
I have a question on this.

If 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 2^9

Then does If 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9 = 2^10?

I think that is how I see it using both the Profs and X and Y's approach.
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30 May 2006, 06:47
OA is A. Great expanations guys. Thanks.
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03 Jun 2006, 15:24
Late A for me same approach as proffs
03 Jun 2006, 15:24
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