Solution to a Contest Math Problem

Michael Hartl • Dec 20, 2023

In Personal Goals for Math Learning, I mentioned that contest math is a great way to practice problem-solving skills while also reviewing the high-school curriculum from a challenging perspective. Among other things, I mentioned a particular problem (together with a hint) that shows just how challenging “high school” mathematics can be in the context of contest math. This post contains a full solution to that problem, as well as a little discussion and a straightforward generalization.

1 A clever manipulation

The problem itself appears in The Art of Problem Solving, Volume 1: The Basics, which is one of the resources for contest math mentioned in the post Resources for Math Learning (Intro to Math). In particular, it is Exercise 7-2 from Chapter 7 (“Special Factorizations and Clever Manipulations”):

Find the sum

\begin{equation} \label{eq:aops_cubes} \frac{1}{\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{4}} + \frac{1}{\sqrt[3]{4} + \sqrt[3]{6} + \sqrt[3]{9}} + \frac{1}{\sqrt[3]{9} + \sqrt[3]{12} + \sqrt[3]{16}}. \end{equation}

You might be tempted to try something like rationalizing the denominators, and indeed this is exactly what’s needed—but how? Well, the chapter name strongly suggests that a mere “straightforward” manipulation probably won’t be enough—it’s going to take something “clever”.

A clue to the solution comes from the chapter context, which includes formulas for the difference and sum of cubes:

\begin{equation} \label{eq:difference_of_cubes} a^3 - b^3 = (a - b)(a^2 + ab + b^2) \end{equation}

and

\begin{equation} \label{eq:sum_of_cubes} a^3 + b^3 = (a + b)(a^2 - ab + b^2). \end{equation}

To see why this is relevant, observe that the denominators in Eq. (1) can be written as

\begin{align*} \sqrt[3]{1^2} + \sqrt[3]{1\cdot2} + \sqrt[3]{2^2} \\ \sqrt[3]{2^2} + \sqrt[3]{2\cdot3} + \sqrt[3]{3^2} \\ \sqrt[3]{3^2} + \sqrt[3]{3\cdot4} + \sqrt[3]{4^2}. \end{align*}

These are of the form

\[ \sqrt[3]{n^2} + \sqrt[3]{n(n + 1)} + \sqrt[3]{(n + 1)^2}, \]

where \( n \) runs from \( 1 \) to \( 3 \), and can be rewritten as

\begin{equation} \label{eq:sum_of_cube_roots} \left(\sqrt[3]{n}\right)^2 + \sqrt[3]{n}\sqrt[3]{(n + 1)} + \left(\sqrt[3]{(n + 1)}\right)^2. \end{equation}

With the definitions \( a = \sqrt[3]{n} \) and \( b = \sqrt[3]{n + 1} \), Eq. (4) becomes

\begin{equation} \label{eq:a_b_form} a^2 + ab + b^2, \end{equation}

which is suggestive because it’s one of the terms in the expression for \( a^3 - b^3 \) shown in Eq. (2). Moreover, in the present case \( a^3 - b^3 \) takes an especially simple form:

\begin{equation} \label{eq:a3_m_b3} a^3 - b^3 = (\sqrt[3]{n})^3 - (\sqrt[3]{n + 1})^3 = n - (n + 1) = -1. \end{equation}

This is what will allow us to rationalize the denominators and thus solve the problem.

2 Simplifying the original expression

Looking back at Eq. (1), we see that it can be written as

\[ \sum_{n=1}^3 \frac{1}{\sqrt[3]{n^2} + \sqrt[3]{n(n + 1)} + \sqrt[3]{(n + 1)^2}}. \]

Eq. (4) and Eq. (5) show that each term in the sum can be written in terms of \( a \) and \( b \):

\[ \frac{1}{a^2 + ab + b^2}. \]

Multiplying top and bottom by \( a - b \) and using Eq. (2) and Eq. (6) then gives

\begin{align*} \frac{1}{a^2 + ab + b^2}\cdot\frac{a-b}{a-b} & = \frac{a-b}{(a-b)(a^2 + ab + b^2)} \\ & = \frac{a-b}{a^3 - b^3} & & \text{Eq. \eqref{eq:difference_of_cubes}} \\ & = \frac{a-b}{-1} & & \text{Eq. \eqref{eq:a3_m_b3}} \\ & = -a + b \\ & = -\sqrt[3]{n} + \sqrt[3]{n + 1}, \end{align*}

where in the final line we have restored the use of the original variable \( n \).

In other words, we have shown that each term in the original expression Eq. (1) can be simplified as follows:

\[ \frac{1}{\sqrt[3]{n^2} + \sqrt[3]{n(n + 1)} + \sqrt[3]{(n + 1)^2}} = -\sqrt[3]{n} + \sqrt[3]{n + 1} . \]

The full sum can thus be written as

\begin{equation} \label{eq:full_sum} \sum_{n=1}^3 (-\sqrt[3]{n} + \sqrt[3]{n + 1}). \end{equation}

3 A wild telescope appears

images/figures/collapsing_telescope
Figure 1: A collapsing telescope.

Eq. (7) is a special kind of expression known as telescoping sum, so named because the sum “collapses” like an officer’s spyglass telescope (Figure 1)1. In particular, writing out the sum term by term shows that all but the first and last terms cancel pairwise, yielding a much simpler expression at the end:

\begin{align*} \sum_{n=1}^3 (-\sqrt[3]{n} + \sqrt[3]{n + 1}) & = (-\sqrt[3]{1} + \sqrt[3]{1 + 1}) + (-\sqrt[3]{2} + \sqrt[3]{2 + 1}) + (-\sqrt[3]{3} + \sqrt[3]{3 + 1}) \\ & = (-\sqrt[3]{1} + \sqrt[3]{2}) + (-\sqrt[3]{2} + \sqrt[3]{3}) + (-\sqrt[3]{3} + \sqrt[3]{4}) \\ & = -\sqrt[3]{1} + (\hl{\sqrt[3]{2} - \sqrt[3]{2}}) + (\hl{\sqrt[3]{3} - \sqrt[3]{3}}) + \sqrt[3]{4} \\ & = -\sqrt[3]{1} + \sqrt[3]{4}. \end{align*}

Noting the further simplification \( \sqrt[3]{1} = 1 \) and rearranging the order to put the positive term first then yields the final answer:

\[ \sqrt[3]{4} - 1 \approx 0.587. \]

4 A straightforward generalization

In a sense, the original expression in Eq. (1) makes the problem harder than it would be if we were asked instead to evaluate a more general sum with arbitrary limits \( k \) and \( m \):

\[ \sum_{n=k}^m \frac{1}{\sqrt[3]{n^2} + \sqrt[3]{n(n + 1)} + \sqrt[3]{(n + 1)^2}}. \]

In this case, we would be much less tempted to try straightforward manipulations since the sum is so complicated that there must be a trick. And indeed the same expression Eq. (7) applies with \( k \) in place of \( 1 \) and \( m \) in place of \( 3 \), so the same pairwise calculation occurs, yielding as before a much simpler expression:

\begin{align*} & \sum_{n=k}^m \frac{1}{\sqrt[3]{n^2} + \sqrt[3]{n(n + 1)} + \sqrt[3]{(n + 1)^2}} \\ & = \sum_{n=k}^m (-\sqrt[3]{n} + \sqrt[3]{n + 1}) \\ & = (-\sqrt[3]{k} + \sqrt[3]{k + 1}) + (-\sqrt[3]{k + 1} + \sqrt[3]{k + 2}) + \cdots \\ & \qquad + (-\sqrt[3]{m - 1} + \sqrt[3]{m}) + (-\sqrt[3]{m} + \sqrt[3]{m + 1}) \\ & = -\sqrt[3]{k} + (\hl{\sqrt[3]{k + 1} -\sqrt[3]{k + 1}}) + \sqrt[3]{k + 2} + \cdots \\ & \qquad - \sqrt[3]{m - 1} + (\hl{\sqrt[3]{m} -\sqrt[3]{m}}) + \sqrt[3]{m + 1} \\ & = -\sqrt[3]{k} + \sqrt[3]{m + 1} \\ & = \sqrt[3]{m + 1} - \sqrt[3]{k}. \end{align*}

We can recover our solution to the original problem by setting \( k = 1 \) and \( m = 3 \), yielding \( \sqrt[3]{4} - \sqrt[3]{1} = \sqrt[3]{4} - 1 \) as required. \( \qed \)

1. Image downloaded on 2023-12-15 and used under the terms of the Creative Commons Attribution-Share Alike 3.0 Unported license. The image has been rotated and flipped, so per the share-alike requirement the modified image is hereby released under the same license.