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M17 q5

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M17 q5 [#permalink]

28 Jul 2012, 02:38

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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}

I used geometric series approach to solve this one. I got an answer like 1/2( 1-(1/2)^21)/(1-1/2) using formula for sum of finite geometric series. This value is very close to one (almost one). How to decide between C Vs D here ..I am confused.

Is there any other way to solve?

[Reveal] Spoiler: OA

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Re: M17 q5 [#permalink]

28 Jul 2012, 02:48

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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Re: M17 q5 [#permalink]

28 Jul 2012, 03:41

teal wrote:

The given sum is
\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\frac{1}{1024^2}.
I deliberately wrote the term 2^{20} as 1024^2, to emphasize the size of it.

Try to think visually:
Take a line segment of length 1, with endpoints 0 and 1. \frac{1}{2} is the length of the left half segment, to which you have to add the half of the remaining
right segment of length \frac{1}{4}(=\frac{1}{2}*\frac{1}{2}). And you continue to add all the time the half of the remaining segment to the right. You do it in fact 19 times. I hope you get the feeling that after all the above steps, you get very, very close to 1, as what is missing from your "chopped-up" line segment is a tiny bit of length \frac{1}{2^{20}}. (See the attached drawing.)

The sum is less than 1 and what is missing is much smaller than 1/10, so the sum must definitely be larger than 9/10.

Therefore, Answer D.

Attachments

GeomSeries.jpg [ 6.55 KiB | Viewed 338 times ]

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Re: M17 q5 [#permalink] 28 Jul 2012, 03:41

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