Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 17 Sep 2014, 12:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M17

Author Message
Senior Manager
Joined: 05 Oct 2008
Posts: 273
Followers: 3

Kudos [?]: 44 [0], given: 22

M17 [#permalink]  02 Oct 2009, 23:31
What is \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20} between?

* \frac{1}{2} and \frac{2}{3}
* \frac{2}{3} and \frac{3}{4}
* \frac{3}{4} and \frac{9}{10}
* \frac{9}{10} and \frac{10}{9}
* \frac{10}{9} and \frac{3}{2}
Math Expert
Joined: 02 Sep 2009
Posts: 29670
Followers: 3494

Kudos [?]: 26278 [0], given: 2708

Re: M17 [#permalink]  03 Oct 2009, 06:39
Expert's post
study 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?

* \frac{1}{2} and \frac{2}{3}
* \frac{2}{3} and \frac{3}{4}
* \frac{3}{4} and \frac{9}{10}
* \frac{9}{10} and \frac{10}{9}
* \frac{10}{9} and \frac{3}{2}

We have geometric progression with:
b_1=\frac{1}{2}, q=\frac{1}{2} and n=20;

S_n=\frac{b_1(1-q^n)}{(1-q)};
S_{20}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{(1-\frac{1}{2})}=1-\frac{1}{2^{20}}, clearly the value of 1-\frac{1}{2^{20}} is less than 1, \frac{1}{2^{20}} is less than \frac{1}{10}, so 1-\frac{1}{2^{20}} will be between \frac{9}{10} and \frac{10}{9}.

Generally speaking when we have geometric progression with common difference in the range -1<q<1 and n is + infinite then the sum of the terms is given by: Sum=\frac{b_1}{(1-q)}.

So, in our case sum tends to become Sum=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 as n increases. Which means that the sum of this sequence never will 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.
_________________
Intern
Joined: 07 Dec 2009
Posts: 15
Followers: 0

Kudos [?]: 3 [0], given: 4

Re: M17 [#permalink]  18 Dec 2009, 21:25
How did you get this part "1/2^20 is less than 1/10" ? And how do you end up leading up to this part "9/10 and 10/9"? I understand that 1-1/2^20 is less than 1 for sure hence 9/10 but how did you get 10/9? A detailed explanation will be appreciated if you have time. Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 29670
Followers: 3494

Kudos [?]: 26278 [1] , given: 2708

Re: M17 [#permalink]  19 Dec 2009, 14:22
1
KUDOS
Expert's post
melissawlim wrote:
How did you get this part "1/2^20 is less than 1/10" ? And how do you end up leading up to this part "9/10 and 10/9"? I understand that 1-1/2^20 is less than 1 for sure hence 9/10 but how did you get 10/9? A detailed explanation will be appreciated if you have time. Thanks!

First of all: when we have geometric progression with common ratio q in the range -1<q<1 (|q|<1), then the sum of the progression: b_1, b_2, ...b_n...b_{+infinity} is Sum=\frac{b_1}{1-q}.

In our case b_1=\frac{1}{2} and q=\frac{1}{2}<1. The sum of the sequence \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}+...+\frac{1}{2^{infinity}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1. Which means that the sum of sequence given, even if it'll continue endlessly, will NEVER be more than 1 (actually it'll be equal to 1 as we calculated).

We have n=20, which is big enough to conclude that the sum will be very close to 1, but again never more than 1.

Another way:

We have geometric progression with:
b_1=\frac{1}{2}, q=\frac{1}{2}, n=20.

Sum=\frac{b_1(1-q^n)}{1-q}

S_{20}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{1-\frac{1}{2}}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{\frac{1}{2}}=1-\frac{1}{2^{20}}

Now: \frac{1}{2^{20}} is less than \frac{1}{10}. Why? \frac{1}{2^4}=\frac{1}{16}<\frac{1}{10}, so if \frac{1}{2^4} is less than \frac{1}{10}, \frac{1}{2^{20}} will be much less than \frac{1}{10}.

Next if we subtract the value less than \frac{1}{10} from 1, we'll get the value more than \frac{9}{10}, as 1-\frac{1}{10}=\frac{9}{10}

Hence 1-\frac{1}{2^{20}} is more than 9/10 and clearly less than 1. The sum is between 9/10 and 1, only answer choice covering this range is D.

Hope it's clear.
_________________
Manager
Joined: 25 Aug 2009
Posts: 177
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 8

Kudos [?]: 99 [0], given: 3

Re: M17 [#permalink]  19 Dec 2009, 16:53
Same approach got D.
_________________

Rock On

Intern
Joined: 07 Dec 2009
Posts: 15
Followers: 0

Kudos [?]: 3 [0], given: 4

Re: M17 [#permalink]  19 Dec 2009, 18:42
Bunuel wrote:
melissawlim wrote:
How did you get this part "1/2^20 is less than 1/10" ? And how do you end up leading up to this part "9/10 and 10/9"? I understand that 1-1/2^20 is less than 1 for sure hence 9/10 but how did you get 10/9? A detailed explanation will be appreciated if you have time. Thanks!

First of all: when we have geometric progression with common ratio q in the range -1<q<1 (|q|<1), then the sum of the progression: b_1, b_2, ...b_n...b_{+infinity} is Sum=\frac{b_1}{1-q}.

In our case b_1=\frac{1}{2} and q=\frac{1}{2}<1. The sum of the sequence \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}+...+\frac{1}{2^{infinity}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1. Which means that the sum of sequence given, even if it'll continue endlessly, will NEVER be more than 1 (actually it'll be equal to 1 as we calculated).

We have n=20, which is big enough to conclude that the sum will be very close to 1, but again never more than 1.

Another way:

We have geometric progression with:
b_1=\frac{1}{2}, q=\frac{1}{2}, n=20.

Sum=\frac{b_1(1-q^n)}{1-q}

S_{20}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{1-\frac{1}{2}}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{\frac{1}{2}}=1-\frac{1}{2^{20}}

Now: \frac{1}{2^{20}} is less than \frac{1}{10}. Why? \frac{1}{2^4}=\frac{1}{16}<\frac{1}{10}, so if \frac{1}{2^4} is less than \frac{1}{10}, \frac{1}{2^{20}} will be much less than \frac{1}{10}.

Next if we subtract the value less than \frac{1}{10} from 1, we'll get the value more than \frac{9}{10}, as 1-\frac{1}{10}=\frac{9}{10}

Hence 1-\frac{1}{2^{20}} is more than 9/10 and clearly less than 1. The sum is between 9/10 and 1, only answer choice covering this range is D.

Hope it's clear.

Thanks, that is very clear. Kudos!
Senior Manager
Joined: 30 Aug 2009
Posts: 290
Location: India
Concentration: General Management
Followers: 3

Kudos [?]: 95 [0], given: 5

Re: Progression [#permalink]  29 Dec 2009, 04:33
study wrote:
How do we use the arithmetic or geometric progression in this particula problem? Anyone?

What is \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20} between?

* \frac{1}{2} and \frac{2}{3}
* \frac{2}{3} and \frac{3}{4}
* \frac{3}{4} and \frac{9}{10}
* \frac{9}{10} and \frac{10}{9}
* \frac{10}{9} and \frac{3}{2}

this will be a GP series with a[1st term] = 1/2 and r[common ratio] =1/2 so
Sum of this series will be a(1-r^n)/(1-r) = 1/2 (1-[1/2]^20]/(1-1/2)= 1 - 1/(2^20)

from the given options option 4 is the best fit
Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 76

Kudos [?]: 480 [0], given: 25

Re: s03#2, Didn't get the explanation, anyone? [#permalink]  03 Nov 2010, 00:24
AtifS 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}

This is from GMAT Club Tests and I didn't get the explanation. Can any expert help with the explanation of this question?

Fact 1 : Sum of infinite series (1/2) + (1/2)^2 + (1/2)^3 + ..... to infinity is 1. (Using infinite GP formula). So option E is out.

Next thing is how does this sum look like ?
1 term : 0.5
2 terms : 1/2 + 1/4 = 0.75
3 terms : 1/2 + 1/4 + 1/8 = 7/8 = 0.875
... and so on

Basically with every term, the distance to 1, is halfed.

Option A can't be right either as it is bounded by 2/3, which is lower than the first 2 terms
Option B can't be right either as it is bounded by 3/4, which is lower than the first 3 terms

Choosing between C & D
This is where it gets interesting
As I mentioned, with each term, the distance to 1 is halfed.
After 2 terms we are 1/4 away
After 3 terms we are 1/8 away
...
After 20 terms we will be 1/(2^20 ) away

This is a very small number, much smaller than 0.1 (which is the bound implied by C)

Hence, the answer must be D

Direct Approach for the question

You can always use the sum of a GP formula, which will immediately give this sum to be equal to 1-\frac{1}{2^{20}}, which then makes it easy to pick D

_________________
Intern
Joined: 02 Nov 2010
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: s03#2, Didn't get the explanation, anyone? [#permalink]  03 Nov 2010, 00:34
You need to know the formula for a geometric series:

SUM(ar^k) = \frac{a(r-r^{n+1})}{1-r} where k = m to n

For this series a = 1, k = \frac{1}{2} and n=20

you'll get the formula \frac{\frac{1}{2}-\frac{1}{2^{21}}}{\frac{1}{2}}, approximate \frac{1}{2^{20}} with zero and you'll get 1. So you know that the series converges towards 1,

Add the first 4 terms to get 15/16 which is bigger than 9/10

Last edited by Papperlapub on 03 Nov 2010, 02:00, edited 3 times in total.
Senior Manager
Joined: 20 Jan 2010
Posts: 278
Schools: HBS, Stanford, Haas, Ross, Cornell, LBS, INSEAD, Oxford, IESE/IE
Followers: 14

Kudos [?]: 138 [0], given: 117

Re: Progression [#permalink]  03 Nov 2010, 01:14
Oops! my bad for posting the already asked question. I searched for it but didn't find in results. I think, I was searching the word s03, which showed only s03#1 (& s03#3) and didn't think of searching for progression (silly me). I think, introducing the new tag "Source:GMAT Club Tests" would be great.

Also, I did apply GP the way you guys mentioned but I think I didn't get the question very well. Is this question asking about where does the sum of the terms in question exist between given answer options? Am I getting it right?
_________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so."
Target=780
http://challengemba.blogspot.com
Kudos??

Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 76

Kudos [?]: 480 [0], given: 25

Re: Progression [#permalink]  04 Nov 2010, 00:22
Yes, it is asking for the range in which the sum sits amongst the given ranges
_________________
Senior Manager
Joined: 23 Oct 2010
Posts: 384
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 12

Kudos [?]: 135 [0], given: 73

Re: M17 [#permalink]  24 Jun 2013, 08:51
is there any other way (such as worst/best scenario) to solve this question?
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: M17   [#permalink] 24 Jun 2013, 08:51
Similar topics Replies Last post
Similar
Topics:
M17#16 2 07 Oct 2009, 20:13
m 17 #27 9 27 Jun 2009, 12:05
18 m17 #8 19 05 Dec 2008, 18:08
8 M17Q5 20 30 Oct 2008, 17:05
1 M17-16 17 26 May 2008, 02:55
Display posts from previous: Sort by

# M17

Moderators: Bunuel, WoundedTiger

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.