Did not find this in a previous thread. Sorry if it was.
2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)
2^9
2^10
2^16
2^35
2^37
What is the fastest way to do this problem?
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GPREP 2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)
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quriousaddict
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Feep
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Anyone familiar with binary will have absolutely no problem here, but if you're not, think about this:
Let's call 2^8 "x". Therefore, 2^7 = 0.5x, 2^6 = 0.25x, 2^5 = 0.125x, and so on, down the line. Adding these up, you get progressively closer to 2x, and then finally, the 2 + 2 seals it.
2^9.
Let's call 2^8 "x". Therefore, 2^7 = 0.5x, 2^6 = 0.25x, 2^5 = 0.125x, and so on, down the line. Adding these up, you get progressively closer to 2x, and then finally, the 2 + 2 seals it.
2^9.
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quriousaddict
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Feep, thanks for your explanation. could you explain how the 2+2 or (x/2^7)+(x/2^7) seals it? Is the point of the problem to see that the equation adds close to 2x? Thanks
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Feep
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Well, if at any point you were allowed to add the same step twice, you would finish the sequence to 2x. 2^8 + 2^7 + 2^6 + 2^5 + (let's stop here and finish it up) 2^5 would be x + x/2 + x/4 + x/8 + x/8.
But this isn't really important. You can see that the sum is getting closer and closer to 2x, not 4x or 8x or whatever. You don't really need to think about those final little tiny parts, because there's no way they could skew the answer so heavily.
But this isn't really important. You can see that the sum is getting closer and closer to 2x, not 4x or 8x or whatever. You don't really need to think about those final little tiny parts, because there's no way they could skew the answer so heavily.
I tutor GMAT/GRE level mathematics privately in the Los Angeles region, as well as via Skype for a discounted rate. Send me a message if you're interested.
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sureshbala
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This can be answered quickly if you are aware of the formula for the Sum of the n terms in a G.P
Here is a general GP
a, ar, ar^2, ar^3, .....ar^(n-1)
Sum to n terms in a G.P = a(r^n - 1)/(r - 1) ......(if r > 1)
= a(1 - r^n)/(1 - r) ......(if r < 1)
Given 2 + 2 + 2^2 + .............+2^8
= 2 + (2 + 2^2 + 2^3+ .........+2^8) (the series in the braces is in G.P with 8 terms)
= 2 + 2(2^8 - 1)/(2 - 1) = 2 + 2(2^8 -1) = 2 + 2^9 - 2 = 2^9
Here is a general GP
a, ar, ar^2, ar^3, .....ar^(n-1)
Sum to n terms in a G.P = a(r^n - 1)/(r - 1) ......(if r > 1)
= a(1 - r^n)/(1 - r) ......(if r < 1)
Given 2 + 2 + 2^2 + .............+2^8
= 2 + (2 + 2^2 + 2^3+ .........+2^8) (the series in the braces is in G.P with 8 terms)
= 2 + 2(2^8 - 1)/(2 - 1) = 2 + 2(2^8 -1) = 2 + 2^9 - 2 = 2^9
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skang357
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If you look at the progression each number after the first is the sum of all of the previous numbers...quriousaddict wrote:Did not find this in a previous thread. Sorry if it was.
2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)
2^9
2^10
2^16
2^35
2^37
What is the fastest way to do this problem?
2 = 2
2+2 = 2^2
2 + 2 + 2^2 = 2^3
2 + 2 + 2^2 + 2^3 = 2^4
So you can see the pattern...
so the next in the series is 2^9, which is the sum of 2+2 + 2^2 + 2^3 .... 2^8
I know this question comes out early in the GMATprep.
I believe this is a 600 level question so I don't think a problem like this requires you to know anything very intuitive such as figuring out that all of the numbers add up to 2X eventually....
Impossible is nothing
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lunarpower
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there are several ways.
(1) PATTERN RECOGNITION
it should be clear that there's nothing special about 2^8 as an ending point; in other words, they just cut the sequence off at a random point. therefore, if we investigate smaller "versions" of the sequence, we should be able to detect a pattern.
let's look:
first term = 2
sum of first 2 terms = 4
sum of first 3 terms = 8
sum of first 4 terms = 16
ok, it's clear what's going on: each new term doubles the sum. if you see a pattern this clear, it doesn't batter whether you understand WHY the pattern exists; just continue it.
so, i want the sum of nine terms, so i'll just double the sum five more times:
32, 64, 128, 256, 512.
this is choice (a).
this is a general rule, by the way: IF SOMETHING CONTAINS MORE THAN 4-5 IDENTICAL STEPS, YOU SHOULD BE ABLE TO EXTRACT A PATTERN FROM LOOKING AT SIMILAR EXAMPLES WITH FEWER STEPS.
(2) ALGEBRA WITH EXPONENTS ("textbook method")
the first two terms are 2 + 2. this is 2(2), or 2^2.
now, using this combined term as the "first term", the first two terms are 2^2 + 2^2. this is 2(2^2), or (2^1)(2^2), or 2^3.
now, using this combined term as the "first term", the first two terms are 2^3 + 2^3. this is 2(2^3), or (2^1)(2^3), or 2^4.
you can see that this will keep happening, so it will continue all the way up to 2^8 + 2^8, which is 2(2^8) = (2^1)(2^8) = 2^9.
(3) ESTIMATE
these answer choices are ridiculously far apart, so you should be able to estimate the answer.
memorize some select powers of 2. notably, 2^10 = 1024, which is "about 1000". 2^9 = 512, which is "about 500". and of course you should know all the smaller ones (2^6 and below) by heart.
thus we have 2^8 is about 250, and the other terms are 128, 64, 32, 16, 8, 4, 2, 2.
looking at these numbers, i'd make a ROUGH ESTIMATE WITHIN A FEW SECONDS:
250 is 250.
128 is ~130.
64 and 32 together are ~100.
the others look like thirty or so together.
so, 250 + 130 + 100 + 30 = 510.
the only answer choice within shouting range is (a); the others are absurdly huge.
--
even if you have no idea how to do anything else, you should still be able to do out the arithmetic within the two-minute time limit.
it won't be fun, but you should be able to do it. if you can't, then the reason is probably "you stared at the problem for too long, and didn't get started when you should have".
(1) PATTERN RECOGNITION
it should be clear that there's nothing special about 2^8 as an ending point; in other words, they just cut the sequence off at a random point. therefore, if we investigate smaller "versions" of the sequence, we should be able to detect a pattern.
let's look:
first term = 2
sum of first 2 terms = 4
sum of first 3 terms = 8
sum of first 4 terms = 16
ok, it's clear what's going on: each new term doubles the sum. if you see a pattern this clear, it doesn't batter whether you understand WHY the pattern exists; just continue it.
so, i want the sum of nine terms, so i'll just double the sum five more times:
32, 64, 128, 256, 512.
this is choice (a).
this is a general rule, by the way: IF SOMETHING CONTAINS MORE THAN 4-5 IDENTICAL STEPS, YOU SHOULD BE ABLE TO EXTRACT A PATTERN FROM LOOKING AT SIMILAR EXAMPLES WITH FEWER STEPS.
(2) ALGEBRA WITH EXPONENTS ("textbook method")
the first two terms are 2 + 2. this is 2(2), or 2^2.
now, using this combined term as the "first term", the first two terms are 2^2 + 2^2. this is 2(2^2), or (2^1)(2^2), or 2^3.
now, using this combined term as the "first term", the first two terms are 2^3 + 2^3. this is 2(2^3), or (2^1)(2^3), or 2^4.
you can see that this will keep happening, so it will continue all the way up to 2^8 + 2^8, which is 2(2^8) = (2^1)(2^8) = 2^9.
(3) ESTIMATE
these answer choices are ridiculously far apart, so you should be able to estimate the answer.
memorize some select powers of 2. notably, 2^10 = 1024, which is "about 1000". 2^9 = 512, which is "about 500". and of course you should know all the smaller ones (2^6 and below) by heart.
thus we have 2^8 is about 250, and the other terms are 128, 64, 32, 16, 8, 4, 2, 2.
looking at these numbers, i'd make a ROUGH ESTIMATE WITHIN A FEW SECONDS:
250 is 250.
128 is ~130.
64 and 32 together are ~100.
the others look like thirty or so together.
so, 250 + 130 + 100 + 30 = 510.
the only answer choice within shouting range is (a); the others are absurdly huge.
--
even if you have no idea how to do anything else, you should still be able to do out the arithmetic within the two-minute time limit.
it won't be fun, but you should be able to do it. if you can't, then the reason is probably "you stared at the problem for too long, and didn't get started when you should have".
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
Approach this problem like you would approach any problem with exponential terms with the same base -- factor it.
Note that every term is a power of 2. You can factor out one 2 from the sum, but that doesn't buy us much. If we combine 2 + 2 = 2*2 = 2^2, we can now factor out a factor of 2^2.
After combining 2 + 2 we get the following: 2^2 + 2^2 + 2^3 + 2^4 + ... + 2^8.
But we can combine 2^2 + 2^2 as we did above. 2^2+2^2 = 2*2^2 = 2^3. Now we have: 2^3 + 2^3 + 2^4 + ... + 2^8.
Combine 2^3 +2^3 as we did above: 2^3 + 2^3 = 2*2^3 = 2^4.
Continue this process and we get 2^8 + 2^8 = 2*2^8 = 2^9.
I think this is the most intuitive approach.
Note that every term is a power of 2. You can factor out one 2 from the sum, but that doesn't buy us much. If we combine 2 + 2 = 2*2 = 2^2, we can now factor out a factor of 2^2.
After combining 2 + 2 we get the following: 2^2 + 2^2 + 2^3 + 2^4 + ... + 2^8.
But we can combine 2^2 + 2^2 as we did above. 2^2+2^2 = 2*2^2 = 2^3. Now we have: 2^3 + 2^3 + 2^4 + ... + 2^8.
Combine 2^3 +2^3 as we did above: 2^3 + 2^3 = 2*2^3 = 2^4.
Continue this process and we get 2^8 + 2^8 = 2*2^8 = 2^9.
I think this is the most intuitive approach.
