# [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
5 years, 9 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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.

Staff - 5 years, 9 months ago

It's a great tricky solution. I like it.

- 5 years, 9 months ago

Solution A and Solution B are correct. Solution C does not show that why $$\displaystyle \sum_{n=3}^{16}\frac{(n-3)!}{n!}=\frac{16^2-16-2}{4\times16(16-1)}$$.

- 5 years, 9 months ago

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.

- 5 years, 9 months ago

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?

- 5 years, 9 months ago

Yes it is, thank you.

- 5 years, 9 months ago

Correct me if I made any errors.

- 5 years, 9 months ago

Count me in. I cant understand C either.

- 5 years, 9 months ago

Staff - 5 years, 9 months ago

I think solution B is almost same as solution A.

- 5 years, 9 months ago

Solution B is more good than A . Can't get solution C

- 5 years, 9 months ago

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.

Staff - 5 years, 9 months ago

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 $$a + b = 119+480 = 599$$

Staff - 5 years, 9 months ago

Unclear juxtaposition.

- 5 years, 9 months ago

Solution C - $$S = \frac {1}{1\times 2 \times 3} + \ldots + \frac {1} {(n)(n+1)(n+2)} + \ldots +\frac {1}{14 \times (14+1) \times (14+2)}$$ $$= \displaystyle \sum_{n=3}^{16} \frac {(n-3)!}{n!} =$$ $$\frac {16^2 - 16 -2}{4 \times 16(16-1)} = \frac {119}{480} = \frac {a} {b}$$ $$a+b = 599$$

Staff - 5 years, 9 months ago

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.

- 5 years, 9 months ago

Agree

- 5 years, 9 months ago

Can you please explain the solution, it's just as Mirza B. said

- 5 years, 9 months ago

Actually, I already explained it above.

- 5 years, 9 months ago