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exponents

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by resilient » Thu Apr 10, 2008 11:58 pm
2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10=

1. 2^10
2. 2^11
3. 2^18
4. 2^54
5. 2^56


oa is b? what is the shortest way to do this?
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by Neo2000 » Fri Apr 11, 2008 12:05 am
2^n + 2^n = 2(2^n) = 2^(n+1)

Once you know this, you will realise that your answer is [spoiler]2^11[/spoiler]

p.s. 250 posts!!! :D
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by cjiang16 » Sat Apr 12, 2008 4:49 pm
Using estimation, the answer is B.
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by Stuart@KaplanGMAT » Sat Apr 12, 2008 7:20 pm
cjiang16 wrote:Using estimation, the answer is B.
Elimination/estimation is a great method for this question.
1. 2^10
2. 2^11
3. 2^18
4. 2^54
5. 2^56
Well, we know that 2^10 is just one of the bits, so eliminate (1).

We also know that we're adding 11 different terms.

Even if EVERY ONE were 2^10, that would give us 11*(2^10) or approximately 2*2*3*(2^10)

Well, 2*2*3*(2^10) would be somewhere between 2^13 and 2^14. So, choices (3), (4) and (5) are all WAY too big.

Only (2) is left!
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Re: exponents

by manju_ej » Mon Jul 21, 2008 8:23 am
resilient wrote:2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10=

1. 2^10
2. 2^11
3. 2^18
4. 2^54
5. 2^56


The answer is 2^11
Remove the first term (2) and take a look at the remaining portions. They are in the Geometric Progression.
So the formula to find the sum of n terms of a geometric progression is S=b1*(1-r^n/1-r), where b1 is the first term, r is the ratio between the two variables, in this case r=2 (ie, third term divided by second term)
By using this formula, I got S=2046.
Now add the term '2' we removed to this figure(s=2046) which gives you 2048.
We know 2^10 is 1024 so 2^10*2=2048.
So the answer is 2^10*2^1 ,which is equal to 2^11, option B.
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by sudhir3127 » Mon Jul 21, 2008 8:53 am
i wud use goemetric mean to solve the problem..
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by reachac » Mon Jul 21, 2008 9:47 am
I'd rather use sum of GP to solve.
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