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What is \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20} between? (A) \frac{1}{2} and \frac{2}{3}(B) \frac{2}{3} and \frac{3}{4}(C) \frac{3}{4} and \frac{9}{10}(D) \frac{9}{10} and \frac{10}{9}(E) \frac{10}{9} and \frac{3}{2}Source: GMAT Club Tests - hardest GMAT questions Could you please provide an explanation for the correct answer. Thank You.
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ventivish wrote: 1/2 + (1/2)^2 + (1/2)^3+..........(1/2)^20 is between
a. 1/2 and 2/3
b. 2/3 and 3/4
c.3/4 and 9/10
d. 9/10 and 10/9
e. 10/9 and 3/2
Could you please provide an explanation for the correct answer. Thank You. = 1/2 + (1/2)^2 + (1/2)^3+ (1/2)^4 + ............... + (1/2)^20 We noted that every term in this series is half of the previous one. If we add these figures by taking LCM, the neumerator never becomes exactly equal but very close to the denometor. If the series is expanded to such an extent, the sum should be close to 1. so it has to be D.
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I think it's E.
I concur with Tigers logic . But the number keeps increasing and it definetly goes beyond 0.9.
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CIO
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This sum can't exceed 1, so D is the answer, but not E. HG wrote: I think it's E.
I concur with Tigers logic . But the number keeps increasing and it definetly goes beyond 0.9.
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Look at it this way. There is a stick. Cut it into two halves. Then cut the other half into two halves......and so on.....Thus, sum of each of these pieces should be very close to the length of the stick.....that is 1.
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Yep D is the answer
Thanks Guys!
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scthakur wrote: Look at it this way. There is a stick. Cut it into two halves. Then cut the other half into two halves......and so on.....Thus, sum of each of these pieces should be very close to the length of the stick.....that is 1. But isn't it like, a piece of a stick which is cut into 2 pieces + a piece of another stick which is cut into 4 pieces + a piece of stick which is cut into 8 and so on..
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I used a method with a little less finesse -- .5 + .25 + .125 + .0625 + ~.03 + ~.015 + ~.0075 + ~.00375 = ~0.965 After .5^8, the values get so small that they add a negligible amount to the total. No way the remaining values add another 0.10 or more onto the total, so the answer is D.
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Ans : d as u adding the value you reach near to one but never reach to one
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The sum is equal to 1-(1/2)^20 as the sum of rhe first 20 terms of a geometric progression with initial value 1/2 and common ratio 1/2. 2^20=(2^10)^2=1024^2>1000^2=10^6, then 1-(1/2)^20>1-0.1^6=0.999999 and less than 1.
Answer is D.
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alternative solution. just try to add this set: 1/2 , 1/4 and 1/8 and second set of figures: 1/2 , 1/4, 1/8, 1/16 now observe the pattern. or 7/8 and 15/16 respectively. in both cases the response will be (n-1)/n, the more consecuitive figures we add the closer is the total sum to 1. so D.
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IMO : D I haven't seen the answers here. And I'm also here after so much time. Correct me guys if I'm wrong. My approach was: Let us take a small set: 1/2+ 1/4 + 1/8 + 1/16 = 15/16 < 1 ( Little less than 1 actually) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 31/32 < 1 ( Little less than 1 again) : : : The same would go again for n terms. So IMO the value would not be more than 1 but always very close to 1. So D comes closest.
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GMAT TIGER wrote: ventivish wrote: 1/2 + (1/2)^2 + (1/2)^3+..........(1/2)^20 is between
a. 1/2 and 2/3
b. 2/3 and 3/4
c.3/4 and 9/10
d. 9/10 and 10/9
e. 10/9 and 3/2
Could you please provide an explanation for the correct answer. Thank You. = 1/2 + (1/2)^2 + (1/2)^3+ (1/2)^4 + ............... + (1/2)^20 We noted that every term in this series is half of the previous one. If we add these figures by taking LCM, the neumerator never becomes exactly equal but very close to the denometor. If the series is expanded to such an extent, the sum should be close to 1. so it has to be D. Hey, Tiger, can you plz explain the LCM again? thanks.
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It seems like if you were to manually compute the series and add the values you can start to see a pattern that will help determine the answer, however is there a formula or some type of shortcut to get to answer D? Or is simple computation and pattern recognition the best way to solve a problem like this?
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This is Geometric Progression .....Sum is slightly less than 1 but greater than .9
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ventivish wrote: What is \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20} between? (A) \frac{1}{2} and \frac{2}{3}(B) \frac{2}{3} and \frac{3}{4}(C) \frac{3}{4} and \frac{9}{10}(D) \frac{9}{10} and \frac{10}{9}(E) \frac{10}{9} and \frac{3}{2}Source: GMAT Club Tests - hardest GMAT questions Could you please provide an explanation for the correct answer. Thank You. I went by the old-manual method... 0.5 0.25 0.125 0.0625 so the summation would give the value at around 0.9 above and since the next squares would give ngligable amounts...hence the summation would be between 0.9 to 1.11 so went with option D can anybody suggest me if this is the only way or is there any other easier way to solve this problem...i took 1m25s for this problem...since it is easier one, can we have a method where we could solve it within 30s?
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It can`t be D..
The maximum value the in the series (1/2) + (1/2)^2 + (1/2)^3 + ........ (infinite series) = a/1-r = 1/2 / (1 - 1/2) = 1.
10/9 = 1.1111111
The value 1.00000000000000000000000000000000001 is not acceptable when compared to 1.
so 1.1111111 is totally out of the question.
The next close answer is C.
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ventivish wrote: What is \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20} between? (A) \frac{1}{2} and \frac{2}{3}(B) \frac{2}{3} and \frac{3}{4}(C) \frac{3}{4} and \frac{9}{10}(D) \frac{9}{10} and \frac{10}{9}(E) \frac{10}{9} and \frac{3}{2}Source: GMAT Club Tests - hardest GMAT questions Could you please provide an explanation for the correct answer. Thank You. The value of \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20} is between? A. \frac{1}{2} and \frac{2}{3}B. \frac{2}{3} and \frac{3}{4}C. \frac{3}{4} and \frac{9}{10}D. \frac{9}{10} and \frac{10}{9}E. \frac{10}{9} and \frac{3}{2}We have the sum of a geometric progression with the first term equal to \frac{1}{2} and the common ratio also equal to \frac{1}{2}. Now, the sum of infinite geometric progression with common ratio |r|<1, is sum=\frac{b}{1-r}, where b is the first term. So, if we had infinite geometric progression instead of just 20 terms then its sum would be Sum=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1. Which means that the sum of this sequence will never exceed 1, also as we have big enough number of terms (20) then the sum will be very close to 1, so we can safely choose answer choice D. Answer: D. One can also use direct formula. We have geometric progression with b=\frac{1}{2}, r=\frac{1}{2} and n=20; S_n=\frac{b(1-r^n)}{(1-r)} --> S_{20}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{(1-\frac{1}{2})}=1-\frac{1}{2^{20}}. Since \frac{1}{2^{20}} is very small number then 1-\frac{1}{2^{20}} will be less than 1 but very close to it. Answer: D.
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D it is:
get the ratio:
between the elements is constant, which means it is a geometric series
r = 0,5 --> each element is half of the previous one
Sum of a limited geometric series:
k = 20 --> upper limit of the limited geometric series
s = a0 * (1 - r^k+1) / (1 - r) = 0.5 * -0.999999523162841796875 / -0.5 = 0.999999523162841796875
so the sum cannot exceed 1.
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There are 2 ways we can approach the soln on this question 1. if you add atleast the first 4 terms of the series you actually exceed 0.9. Also we know that positive powers of any number that is greater than zero but less than 1 (fraction) will always be less than the number itself. This essentially means with increasing positive power the number will get smaller. Thus Summation of all these number can never be greater than 1. 2. This is a Geomeric Progression. Use the summation formula: it leads to the following expression (1- .5^21). This means the result of the sum is less than one but very close to 1. Definitely greater than 0.9 but less than 1. Both these approaches lead to answer choice D. Thanks --------------------- Too self-concious to ask for Kundos.
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