melissawlim wrote:

How did you get this part "1/2^20 is less than 1/10" ? And how do you end up leading up to this part "9/10 and 10/9"? I understand that 1-1/2^20 is less than 1 for sure hence 9/10 but how did you get 10/9? A detailed explanation will be appreciated if you have time. Thanks!

First of all: when we have geometric progression with common ratio \(q\) in the range \(-1<q<1\) \((|q|<1)\), then the sum of the progression: \(b_1, b_2, ...b_n...b_{+infinity}\) is \(Sum=\frac{b_1}{1-q}\).

In our case \(b_1=\frac{1}{2}\) and \(q=\frac{1}{2}<1\). The sum of the sequence \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}+...+\frac{1}{2^{infinity}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1\). Which means that the sum of sequence given, even if it'll continue endlessly, will NEVER be more than 1 (actually it'll be equal to 1 as we calculated).

We have n=20, which is big enough to conclude that the sum will be very close to 1, but again never more than 1.

Answer: D.Another way:

We have geometric progression with:

\(b_1=\frac{1}{2}\), \(q=\frac{1}{2}\), \(n=20\).

\(Sum=\frac{b_1(1-q^n)}{1-q}\)

\(S_{20}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{1-\frac{1}{2}}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{\frac{1}{2}}=1-\frac{1}{2^{20}}\)

Now: \(\frac{1}{2^{20}}\) is less than \(\frac{1}{10}\). Why? \(\frac{1}{2^4}=\frac{1}{16}<\frac{1}{10}\), so if \(\frac{1}{2^4}\) is less than \(\frac{1}{10}\), \(\frac{1}{2^{20}}\) will be much less than \(\frac{1}{10}\).

Next if we subtract the value

less than \(\frac{1}{10}\) from \(1\), we'll get the value more than \(\frac{9}{10}\), as \(1-\frac{1}{10}=\frac{9}{10}\)

Hence \(1-\frac{1}{2^{20}}\) is more than 9/10 and clearly less than \(1\). The sum is between 9/10 and 1, only answer choice covering this range is D.

Answer: D.Hope it's clear.

Thanks, that is very clear. Kudos!