The limiting sum of the infinite series [m][fraction]1/2[/fraction] +

Hello again, Asad. I wanted to outline my approach to the question, in case it differs from your own or that of any onlooker. I typically favor a more intuitive approach than a rigorous mathematical approach. In the question at hand, I started by converting all the fractions to thirty-seconds and adding:

\(\frac{16}{32}+\frac{24}{32}+\frac{20}{32}+\frac{14}{32}+\frac{9}{32}=\frac{83}{32}\)

Now, our next fraction would be in sixty-fourths, per the pattern, and the numerator would be 11. To put this value together with the earlier sum in a meaningful way, I converted the earlier answer to sixty-fourths:

\(\frac{11}{64}+\frac{166}{64}=\frac{177}{64}\)

At this point, we have a decimal value that would lie between 2 and 3, but closer to the 3 end, since

\(64*3=192\)

Looking at the answer choices, we can eliminate (A) since that is already too low. How do we know?

\(64*2+32<177\)

We can also appreciate, without doing more math, that the sum will grow in smaller and smaller increments, and our last term was already about one-sixth. Thus, it appears as though our sum will converge on 3. Still, for the inner fact-checker, we could go on. In terms of sixty-fourths, we have 15 more to add in the numerator (to get from 177 to 192), and that is going to take multiple steps. The next two terms would give us

\(\frac{13}{128}+\frac{15}{256}\)

or, in terms of sixty-fourths, the numerators would be

\(\frac{13}{2}+\frac{15}{4}=\frac{41}{4}\)

Closer still, as we would have just 4.75 more to go to get our numerator to 192. But to be honest, I see no way that, in keeping with this pattern, the numerator will increase to 192 and another 32 (half of 64) to get us up to answer (C). Although I would have selected my answer already, if I had not before, I would feel confident choosing (B), confirming, and moving on to the next challenge.

Thank you for sharing the question. What is the source, if you do not mind my asking?

- Andrew

The source is 'Quantum GMAT'... Already source mentioned in the spoiler. This question has been designed for them who are targeting Q51! Thanks..

Ah, the spoiler that I sometimes cannot get to work. I am not sure if it is a browser issue or a computer issue, but sometimes I click on the spoiler or the OA/OE and get nothing. It worked this time, so I feel foolish. (Oh well.) I would like to hit a 51 in Quant myself. Maybe later this year, once matters settle down, I will put myself to the test and get a Verified Score to post... hopefully with a pair of 51s to show for the effort. Time will tell.

- Andrew

good luck for you so that you can get q51! in the mean time you can see the explanation from the following link (if you need..)

https://www.facebook.com/gmatquantum/vi ... 560222132/

Thank you for the well wishes, as well as for the link. (I am not on Facebook, so I would not have found anything on my own that way.) Kudos from me for all the help.

- Andrew

Thank you for the well wishes, as well as for the link. (I am not on Facebook, so I would not have found anything on my own that way.) Kudos from me for all the help.

- Andrew[/quote]
I think you should not have any facebook ID to watch that video! if you click on this link it'll be automatically started..