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M17 Q05

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Manager
Joined: 14 Dec 2011
Posts: 76

Kudos [?]: 57 [0], given: 77

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17 Oct 2012, 09:41
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 $$\frac{3}{2}$$
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Last edited by Bunuel on 17 Oct 2012, 13:02, edited 1 time in total.
Renamed the topic and edited the question.

Kudos [?]: 57 [0], given: 77

Intern
Joined: 02 Nov 2009
Posts: 42

Kudos [?]: 57 [1], given: 8

Location: India
Concentration: General Management, Technology
GMAT Date: 04-21-2013
GPA: 4
WE: Information Technology (Internet and New Media)

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17 Oct 2012, 11:04
1
KUDOS

1/2+1/4 =3/4 or .75 or (2^2-1)/2^2

1/2+1/4+1/8 =7/8 or .875 or (2^3-1)/2^3

1/2+1/4+1/8+1/16 =15/16 >.9
15/16 is also (2^4-1)/2^4

so adding all the terms we will get (2^20-1)/2^20 and this is <1
and we already know that this is >.9 as the sum of four terms itself is >.9

and the only choice which satisfies this is D
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KPV

Kudos [?]: 57 [1], given: 8

Math Expert
Joined: 02 Sep 2009
Posts: 41891

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17 Oct 2012, 13:02
1
KUDOS
Expert's post
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.

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.

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Kudos [?]: 129068 [1], given: 12194

Director
Joined: 22 Mar 2011
Posts: 610

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WE: Science (Education)

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17 Oct 2012, 10:14
vinnik wrote:
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 $$\frac{3}{2}$$

How can i solve this question by an easier approach ?

Thanks & Regards
Vinni

Imagine you have to walk from 0 to 1. You walk half of the way, which is 1/2, and 1/2 left.
You walk now 1/2 of what remained, which is 1/4 and 1/4 remained.
Then you walk 1/2 of what remained, which is 1/8 and 1/8 remained.
.....
The best way to understand is by drawing a line segment and visualizing the distances.
Hope you got the idea. After 20 such walks, you are awfully close to 1, but left of it.
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PhD in Applied Mathematics
Love GMAT Quant questions and running.

Kudos [?]: 1058 [0], given: 43

Manager
Joined: 14 Dec 2011
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18 Oct 2012, 08:59
Thanks everyone. The methods are just amazing.

Bunuel, your direct formula method is fantastic.

Thanks & Regards
Vinni

Kudos [?]: 57 [0], given: 77

Re: M17 Q05   [#permalink] 18 Oct 2012, 08:59
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M17 Q05

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