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

GeomSeries.jpg [ 6.55 KiB | Viewed 1036 times ]

_________________

PhD in Applied Mathematics

Love GMAT Quant questions and running.