I tried other values, such as
t=3
3/4* 48= 36
t=10
(10/11) * 39= 35 ( the number is decreasing )
So, I’m not getting how @gregmat solve it
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I can’t find the video what is the actual answer to the question? Is it QB is greater?
I think the question is testing the subject of SUM in an arithmetic sequence. So I tried to identify the first and the last term within set S, n, and n50.
n1 = 1 - (1/2) = 1/2
n50 = 1 - (1/50) = 49/50
The total # of n = 50
Then, I tried solving it by SUM of arithmetic sequence formula n/2 (n1 + n50) which is:
(50/2) * (37/25) = 37 which is lesser than 42 therefore answer B.
However, I’m not confident with my answer at all, I hope someone could correct me and point out my mistake.
The answer is A. There are a couple of ways you could go around this. The rigorous (but out of scope!) way would be to convert to an integral:
\sum_{t = 1}^{50} \left(1-\frac{1}{1+t}\right) = 50 - \sum_{t = 1}^{50} \left(\frac{1}{1+t}\right)
Then make use of the Euler’s theorem - we replace \epsilon (that’s a little over a half) with 2/3 for simplicity:
\sum_{t = 1}^{50} \left(\frac{1}{1+t}\right) - \int_{1}^{50} \frac{di}{i} < \frac{2}{3}
\sum_{t = 1}^{50} \left(\frac{1}{1+t}\right) - \int_{1}^{50} \frac{di}{i} < \frac{2}{3}
\sum_{t = 1}^{50} \left(\frac{1}{1+t}\right) - \ln 50 < \frac{2}{3}
With the simplification that \ln 50 < 5,
\sum_{t = 1}^{50} \left(\frac{1}{1+t}\right) < \frac{2}{3} + 5
\sum_{t = 1}^{50} \left(\frac{1}{1+t}\right) < \frac{17}{3}
Plugging it back,
\sum_{t = 1}^{50} \left(1-\frac{1}{1+t}\right) = 50 - \sum_{t = 1}^{50} \left(\frac{1}{1+t}\right) > 50 - \frac{17}{3} > 44 > 42
and hence the answer is A.
The problem is that this is not a solution suitable for the GRE for obvious reasons, and in fact I am not sure of a rigorous alternative either. GregMat’s solution is basically to assume every term to be 6/7 - the reason I am not comfortable with that solution is because he is assuming that
\sum_{i = 7}^{50} \frac{i}{i + 1} - \frac{6}{7} > \sum_{i = 1}^{5} \frac{6}{7} - \frac{i}{i + 1}
You could show that
\sum_{i = 7}^{50} \frac{i}{i + 1} - \frac{6}{7} < 7
and hence try to show that Greg’s relation is true, but it’s a bit more hand-wavy than I like.
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we can’t do this because the difference between two terms is not even. This rule applies only for evenly spaced terms.
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@gregmat if you could help pls
I just used logic to solve it. It wasn’t a foolproof method, but seemed accurate