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[Calvin] Which solution will you feature (6)?

Previous discussion.

Below, we present a problem from the 1/21 Algebra and Number Theory set, along with 3 student submitted solution. You may vote up for the solutions that you think should be featured, and should vote down for those solutions that you think are wrong (voting is anonymous!). Also, feel free to make remarks about these solutions, especially since threading of comments has been introduced :).

This question is an old chestnut, and one of my favorite ones. Sums of fractions \( S = \frac {1}{1\times 2 \times 3} + \frac {1}{2\times 3 \times 4} + \ldots + \frac {1} {(n)(n+1)(n+2)} \) \(+ \ldots +\frac {1}{14 \times (14+1) \times (14+2)} = \frac {a} {b}, \) where \(a\) and \(b\) are positive, coprime integers. What is the value of \( a + b \)?

You may try the problem by clicking on the above link.

All solutions may have LaTeX edits to make the math appear properly. The exposition is presented as is, and has not been edited.

\[ \mbox{Remarks from Calvin} \]

Solution A - This is a good approach to use, where you not only realize how to use the telescoping series, but also find a simple series to use. The explanation is clear to understand, and at a glance others can see exactly what you are doing.

However, it makes a false claim, that "As there are three terms in each denominator, the telescoping series may be constructed by using partial sums that have only two terms in their denominator." This only happens to be the case, because we have \( \frac {1}{n(n+1)(n+2)} = \frac {1}{n} - \frac {1}{n+1} - ( \frac {1}{n+1} - \frac {1}{n+2} ) \). For example, if we consider \( \frac {n-3}{n(n+1)(n+3)} = -\frac {1}{n} + 2 \frac {1}{n+1} - \frac {1}{n+3} \), It can be summed using the telescoping series, but doesn't have a nice 'two terms in denominator' expression that will work directly.

Solution B - This solution, in essence, is the exact same as A. However, due to the way it is written up, it's unclear what is exactly happening. Part of the issue is not initially stating that it is a proof by telescoping series, not defining what \( T(n), V(n)\) are supposed to be (since we are not given the context of a telescoping proof), and furthermore suddenly changing it to \( t(n), v(n)\) (are these supposed to be different variables?).

Solution C - While this is technically correct, it provides no explanation of where that formula comes from, or why it is even true. You should not expect the reader to be able to fill in sudden gaps like this, especially if it constitutes the crux of your solution. At the very least, state how to do it and why it works. Kudos to Zi Song for filling in the gap.
It was also missing a punctuation mark between \( [v(1)-v(15)]/2\) and \(v(1) = 1/2\).

I've chosen to feature the solution by Neelam, which explains how to obtain the partial fraction decomposition, and the pairing of the telescoping series. Students who have received this problem may click on the problem link to view the solution.

Note by Calvin Lin
4 years, 6 months ago

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Solution A - This sum is complicated to do directly, but we may rearrange some terms of each fraction separately. As there are three terms in each denominator, the telescoping series may be constructed by using partial sums that have only two terms in their denominator. Therefore, we are looking for a expression like this: \( \frac {1}{(n)(n+1)(n+2)} = A \left( \frac {1}{(n)(n+1)} - \frac {1}{(n+1)(n+2)} \right), \) where A is a constant number. Solving for every possible n, we may find that: \( A = \frac {1}{2} .\) Returning to the first sum S, using the property above: \( S = \frac{1}{2} \left( \frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{2 \times 3} - \frac{1}{3 \times 4} + ... + \frac{1}{14 \times 15} - \frac{1}{15 \times 16} \right). \) Regrouping the terms: \( S= \frac{1}{2} \left( \frac{1}{1 \times 2} - \frac{1}{15 \times 16} \right) = \frac{119}{480} = \frac{a}{b}. \) As: \( 119 = 7 \times 17 \) and \( 480 = 32 \times 3 \times 5 \), we have that 119 and 480 are coprime integers. Therefore, \( a = 119, b = 480 \) and: \( a+b = 599 \), the solution of this problem. Calvin Lin Staff · 4 years, 6 months ago

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@Calvin Lin It's a great tricky solution. I like it. Rushikesh Jogdand · 4 years, 6 months ago

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Solution A and Solution B are correct. Solution C does not show that why \(\displaystyle \sum_{n=3}^{16}\frac{(n-3)!}{n!}=\frac{16^2-16-2}{4\times16(16-1)}\). Qi Huan Tan · 4 years, 6 months ago

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Would it be possible to ask the author of solution C for some elaboration? Since some people are voting it up, it can't be as arbitrary as it seems. I understand that this request has little to do with the actual result of the selection process. Still, i feel it is a good opportunity for us to learn something new. Hope I'm not too much out of topic here. Andreas Philippidis · 4 years, 6 months ago

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@Andreas Philippidis First, we claim that \(\displaystyle \sum^{n}_{k = 3}\frac{(k - 3)!}{k!} = \frac{(n - 2)(n + 1)}{4n(n - 1)}\). We'll prove by inducting on \(n\). Clearly, the base case is true. Now we establish the inductive step. 

\(\displaystyle \sum^{n + 1}_{k = 3}\frac{(k - 3)!}{k!} = \displaystyle \sum^{n}_{k = 3}\frac{(k - 3)!}{k!} + \frac{(n - 2)!}{(n + 1)!}\) \(= \frac{n^{2} - n - 2}{4n(n - 1)} + \frac{1}{(n - 1)n(n + 1)} = \frac{n^{3} - 3n + 2}{4n(n - 1)(n + 1)}\) \(= \frac{(n - 1)(n - 1)(n + 2)}{4n(n + 1)(n - 1)} = \frac{(n - 1)(n + 2)}{4n(n + 1)}\), establishing the inductive step.

Now, from \(\displaystyle \sum^{14}_{k = 1}\frac{1}{k(k + 1)(k + 2)} = \displaystyle \sum^{16}_{k = 3}\frac{(k - 3)!}{k!}\) \(= \frac{(14)(17)}{4(16)(15)} = \frac{119}{480}\), and we get \(a + b = 119 + 480 = 599\)

By the way, I'm not the author of this solution. :) Is this clear enough? Zi Song Yeoh · 4 years, 6 months ago

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@Zi Song Yeoh Yes it is, thank you. Andreas Philippidis · 4 years, 6 months ago

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@Zi Song Yeoh Correct me if I made any errors. Zi Song Yeoh · 4 years, 6 months ago

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@Andreas Philippidis Count me in. I cant understand C either. Vicky Bro · 4 years, 6 months ago

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Remarks have been added. Calvin Lin Staff · 4 years, 6 months ago

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I think solution B is almost same as solution A. Vicky Bro · 4 years, 6 months ago

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Solution B is more good than A . Can't get solution C Chhavi Kansal · 4 years, 6 months ago

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@Chhavi Kansal If you think that solution B is more good than A, then you should vote it up. If you think that solution C is wrong, then you should vote it down. Calvin Lin Staff · 4 years, 6 months ago

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Solution B - we can observe that nth term is \(1/n(n+1)(n+2)\). Let \(V(n)\) be \(1/n(n+1)\). Then if we do \(v(n) - v(n+1)\), we get \(2/n(n+1)(n+2)\). which is equal to \(2. T(n)\) therefore \(t(n) = [v(n) - v(n+1)]/2\). when we do \(t(1) + t(2) + ........... + t(14)\) we get \([v(1) - v(15)]/2 v(1) = 1/2\) and \(v(15) = 1/240\). and then on further simplification we get the answer as \(119/480\). therefore a=119 and b=480 and \(a + b = 119+480 = 599\) Calvin Lin Staff · 4 years, 6 months ago

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@Calvin Lin Unclear juxtaposition. Tim Ye · 4 years, 6 months ago

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Solution C - \( S = \frac {1}{1\times 2 \times 3} + \ldots + \frac {1} {(n)(n+1)(n+2)} + \ldots +\frac {1}{14 \times (14+1) \times (14+2)} \) \( = \displaystyle \sum_{n=3}^{16} \frac {(n-3)!}{n!} = \) \( \frac {16^2 - 16 -2}{4 \times 16(16-1)} = \frac {119}{480} = \frac {a} {b} \) \(a+b = 599\) Calvin Lin Staff · 4 years, 6 months ago

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@Calvin Lin C is pretty neat solution, but younger kids might have problem with this as it is not fully elaborated. A is better than B is terms of clarity and straight forwardness. Mirza Baig · 4 years, 6 months ago

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@Mirza Baig Agree Zi Song Yeoh · 4 years, 6 months ago

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@Calvin Lin Can you please explain the solution, it's just as Mirza B. said Gurpreet Singh · 4 years, 6 months ago

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@Gurpreet Singh Actually, I already explained it above. Zi Song Yeoh · 4 years, 6 months ago

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