Problem 42 - Discrete Mathematics with Applications

Question

From the binomial expansion \((1+x)^{n}=\sum_{r=0}^{n}\left(\begin{array}{l}n \\\ r\end{array}\right) x^{r},\) it can be shown using calculus that \(n(1+x)^{n-1}=\sum_{r=1}^{n}\left(\begin{array}{c}n \\ r\end{array}\right) r x^{r-1} .\) Using this result, prove each. $$1\left(\begin{array}{l}n \\ 1\end{array}\right)+3\left(\begin{array}{l}n \\\ 3\end{array}\right)+5\left(\begin{array}{l}n \\\ 5\end{array}\right)+\cdots=2\left(\begin{array}{l}n \\\ 2\end{array}\right)+4\left(\begin{array}{l}n \\\ 4\end{array}\right)+6\left(\begin{array}{l}n \\ 6\end{array}\right)+\cdots=n 2^{n-2}$$

Step-by-Step Solution

Answer

To prove the given equation, we first rewrite the given result for a specific value of x, which is x = -1. After simplifying the expressions and separating the summation into odd and even terms, we compare it with the given equation and rewrite it in terms of sums of even and odd integers. By canceling out the common factor, we conclude that the given equation is true: $$\left(\begin{array}{c}{n} \\\ {1}\end{array}\right) 1 + \left(\begin{array}{c}{n} \\\ {3}\end{array}\right) 3 + \dots = \left(\begin{array}{c}{n} \\\ {2}\end{array}\right) 2 + \left(\begin{array}{c}{n} \\\ {4}\end{array}\right) 4 + \dots$$

Step 1: Rewrite the given result

The given result states that \(n(1+x)^{n-1}=\sum_{r=1}^{n}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r x^{n-1}\). To prove the given equation, we need to rewrite this result for a specific value of 'x'. Step 2: Choose x value

Step 2: Choose x value

To work with the given result, If we plug in x = -1, we have \(n(1-1)^{n-1}=\sum_{r=1}^{n}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r (-1)^{n-1}\). Step 3: Simplify expressions

Step 3: Simplify expressions

Now, we will simplify the expressions on both sides of this equation. We get \(0 = \sum_{r=1}^{n}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r (-1)^{n-1}\). Step 4: Separate odd and even terms

Step 4: Separate odd and even terms

We will now separate the summation into odd and even terms and simplify further: For odd r, we have \((-1)^{n-1}\) For even r, we have \((-1)^{n-2}\) So, the equation becomes: \(0 = \sum_{r, odd}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r (-1)^{n-1}+\sum_{r, even}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r (-1)^{n-2}\) Step 5: Simplify and compare

Step 5: Simplify and compare

Let's simplify the equation further by factoring out \((-1)^{n-1}\) and \((-1)^{n-2}\) from the odd and even terms, respectively: \(0 = (-1)^{n-1} \sum_{r, odd}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r + (-1)^{n-2}\sum_{r, even} \left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r\) Comparing with the given equation, we have: \(\sum_{r, odd}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r = \sum_{r, even} \left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r\) Divide both sides by 2: \(\sum_{r, odd}\left(\frac{1}{2}\begin{array}{c}{n} \\\ {r}\end{array}\right) r = \sum_{r, even} \left(\frac{1}{2}\begin{array}{c}{n} \\\ {r}\end{array}\right) r\) Step 6: Rewrite in terms of sums of even and odd integers

Step 6: Rewrite in terms of sums of even and odd integers

Now rewrite the left side as a sum of the form: $$\frac{1}{2}\left(\begin{array}{c}{n} \\\ {1}\end{array}\right) 1 + \frac{1}{2}\left(\begin{array}{c}{n} \\\ {3}\end{array}\right) 3 + \dots$$ And the right side as the sum of the form: $$\frac{1}{2}\left(\begin{array}{c}{n} \\\ {2}\end{array}\right) 2 + \frac{1}{2}\left(\begin{array}{c}{n} \\\ {4}\end{array}\right) 4 + \dots$$ We obtain: $$\frac{1}{2}\left(\begin{array}{c}{n} \\\ {1}\end{array}\right) 1 + \frac{1}{2}\left(\begin{array}{c}{n} \\\ {3}\end{array}\right) 3 + \dots = \frac{1}{2}\left(\begin{array}{c}{n} \\\ {2}\end{array}\right) 2 + \frac{1}{2}\left(\begin{array}{c}{n} \\\ {4}\end{array}\right) 4 + \dots$$ Step 7: Cancel out common factor and conclude

Step 7: Cancel out common factor and conclude

Finally, cancel out the common factor \(\frac{1}{2}\) from both sides of the equation: $$\left(\begin{array}{c}{n} \\\ {1}\end{array}\right) 1 + \left(\begin{array}{c}{n} \\\ {3}\end{array}\right) 3 + \dots = \left(\begin{array}{c}{n} \\\ {2}\end{array}\right) 2 + \left(\begin{array}{c}{n} \\\ {4}\end{array}\right) 4 + \dots$$ This proves the given equation.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics focusing on counting, arrangement, and combination of objects. Understanding combinatorics is crucial for grasping the binomial expansion. In binomial expansion exercises, we often determine how many ways we can choose objects from a larger set without regard to order. An example of this is calculating how many different ways you can select two items from a group of five.

The primary tool in combinatorics used in binomial expansion is factorial notation. For instance, to choose \(r\) elements from \(n\) elements, we use the formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). Here, \(n!\) (read as "n factorial") is the product of all positive integers up to \(n\). Combinatorics thus provides the backbone to solve many algebraic identities by methodically counting possibilities.

Calculus

Calculus is the study of how things change and is used to find rates of change. In the context of binomial expansion, calculus can help prove identities that emerge from expanded binomials.

For example, by differentiating the function \((1+x)^n\), we acquire insight into how changes in \(x\) affect the expansion. If we take the derivative, \(\frac{d}{dx}[(1+x)^n] = n(1+x)^{n-1}\), it reveals the sum derived from the binomial coefficients. This is because differentiation, a core concept in calculus, uncovers relationships between quantities by showing their rates of change, leading to deeper algebraic insights.

Binomial Coefficients

Binomial coefficients are represented as \(\binom{n}{r}\) and indicate the number of ways to choose \(r\) elements from \(n\) elements, much like in combinatorics. These coefficients are pivotal to understanding binomial expansion since they serve as the multipliers for each term in the expansion of a binomial expression.

For instance, in the expansion of \((1+x)^n\), each term takes the form \(\binom{n}{r}x^r\), where \(r\) ranges from 0 to \(n\). This forms the body of Pascal's Triangle, a conceptual and visual tool that helps swiftly identify these coefficients. Pascal's Triangle starts with a 1 at the top, and each subsequent row contains coefficients that are the sum of the two directly above it. This pattern displays the relationship and repetition seen within binomial expansions.

Mathematical Proof

Mathematical proof is a logical process to establish the truth of a mathematical statement. It's a step-by-step demonstration that ensures conclusions are logically sound and verified.

In the given exercise, the proof involves showing equality by manipulating the binomial coefficients using a binomial identity that relates sequences of numbers. This uses concepts from both combinatorics and calculus, and each step ensures the terms on both sides of an equation match.

The use of mathematical proof is foundational in mathematics as it provides certainty. It is a process that verifies whether a statement is not just seemingly true, but unequivocally so, under all conditions stipulated. Thus, mathematical proofs ensure that solutions are reliable and universally accepted.