teal 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}

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?

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

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PhD in Applied Mathematics

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