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# M17

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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 $$\frac{3}{2}$$
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03 Oct 2009, 06:39
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.
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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!
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19 Dec 2009, 14:22
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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.
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19 Dec 2009, 16:53
Same approach got D.
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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!
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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
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Re: s03#2, Didn't get the explanation, anyone? [#permalink]

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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

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Re: s03#2, Didn't get the explanation, anyone? [#permalink]

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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.
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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?
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04 Nov 2010, 00:22
Yes, it is asking for the range in which the sum sits amongst the given ranges
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24 Jun 2013, 08:51
is there any other way (such as worst/best scenario) to solve this question?
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Re: M17   [#permalink] 24 Jun 2013, 08:51
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# M17

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