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M17 03 Oct 2009, 00:31
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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?

* \(\frac{1}{2}\) 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 \(\frac{3}{2}\)

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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?

* \(\frac{1}{2}\) 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 \(\frac{3}{2}\)
Show more

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 \(\frac{10}{9}\).

Answer D.

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.
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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!
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M17 19 Dec 2009, 15:22
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melissawlim

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!
Show more

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.

Answer: D.

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.

Answer: D.

Hope it's clear.
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Same approach got D.
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melissawlim

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.

Answer: D.

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.

Answer: D.

Hope it's clear.
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Thanks, that is very clear. Kudos!
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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 \(\frac{3}{2}\)
Show more

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
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AtifS
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?
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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
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M17 Updated on: 03 Nov 2010, 03:00
Originally posted by Papperlapub on 03 Nov 2010, 01:34.
Last edited by Papperlapub on 03 Nov 2010, 03:00, edited 3 times in total.
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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

This leaves answer D as the correct answer (I think :))
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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?
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Yes, it is asking for the range in which the sum sits amongst the given ranges
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is there any other way (such as worst/best scenario) to solve this question?

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