M17-05

Official Solution:

The value of \(\frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + ... + (\frac{1}{2})^{20}\) 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 \(\frac{3}{2}\)


The value of the given expression can be represented as the sum of a geometric sequence with the first term equal to \(\frac{1}{2}\) and the common ratio also equal to \(\frac{1}{2}\).

For an infinite geometric sequence with a common ratio \(|r| < 1\), the sum can be calculated as \(sum = \frac{b}{1-r}\), where \(b\) is the first term. So, if we had an infinite geometric sequence instead of just 20 terms, its sum would be \(Sum = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1\). This means that the sum of this sequence will never exceed 1. Also, since we have a large enough number of terms (20), the sum will be very close to 1, so we can safely choose answer choice D.

We can also use the direct formula for the sum of a finite geometric sequence:

For a geometric sequence with \(b = \frac{1}{2}\), \(r = \frac{1}{2}\), and \(n = 20\):

\(S_n = \frac{b(1 - r^n)}{(1 - r)}\), so:

\(S_{20} = \frac{\frac{1}{2}(1 - \frac{1}{2^{20}})}{(1 - \frac{1}{2})} = 1 - \frac{1}{2^{20}}\). Since \(\frac{1}{2^{20}}\) is a very small number, \(1 - \frac{1}{2^{20}}\) will be less than 1 but very close to it.


Answer: D