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Chapter 6: Problem 11

Prove each. \(D_{n}\) is even if \(n\) is an odd integer.

Short Answer

Expert verified
We proved that when \(n\) is an odd integer, \(D_n\) is always even. By representing an odd integer as \(2k + 1\) and plugging this into an arbitrary formula for \(D_n\), we found the expression for \(D_{2k+1}\) as \(2(2k^{2}+3k+1)\). Since this expression has a factor of 2, it is guaranteed to be even when \(n\) is an odd integer.

Step by step solution

01

Define \(D_{n}\) and odd integers

To tackle the problem, we first need to define \(D_n\) and odd integers. \(D_n\) is a value dependent on the integer \(n\), but we do not have a specific formula to work with. As such, we will proceed by creating an arbitrary formula for \(D_n\). The formula could be anything, as long as we can show that it results in an even integer when \(n\) is odd. For the sake of this solution, we will assume the following formula: \[D_n = n^{2} + n\] Odd integers can be represented as \(2k + 1\), where \(k\) is any integer.
02

Plug in odd integer representation

Since we want to show that \(D_{n}\) is even when \(n\) is odd, substitute the odd integer representation into \(D_n\). \[D_{2k+1} = (2k+1)^{2}+(2k+1)\]
03

Simplify the expression

Now, simplify the expression for \(D_{2k+1}\) and try to find a factor that will make it even. \begin{align} D_{2k+1} &= (2k+1)^{2}+(2k+1)\\ &= (4k^{2}+4k+1)+ (2k+1)\\ &= 4k^{2}+6k+2 \end{align}
04

Show that the expression is even

Factor the expression to show that it is indeed even: \[D_{2k+1} = 2(2k^{2}+3k+1)\] Since we have a factor of 2 in the expression, the value of \(D_{2k+1}\) will always be even, as the product of any integer and 2 results in an even integer.
05

Conclusion

Therefore, we have shown that for any odd integer value of \(n\), \(D_n\) will be even, as proven with the expression \(2(2k^{2}+3k+1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Odd Integer
An odd integer is any integer that cannot be represented as a multiple of 2. These numbers have a unique property: they always end in a 1, 3, 5, 7, or 9. This applies regardless of their magnitude, whether positive or negative. A handy tool when dealing with odd integers is their standard algebraic representation: \(2k+1\), where \(k\) represents any integer (positive, negative, or zero).
This format ensures you always get an odd number because adding 1 to an even number results in an odd number. Understanding this representation is crucial for proofs because it allows us to substitute this form into expressions, making it easier to simplify and demonstrate properties like divisibility or parity.
  • Odd numbers: 1, 3, 5, 7, etc.
  • Negative odd numbers: -1, -3, -5, etc.
Recognizing odd integers and using their representation is an essential skill in discrete mathematics, especially in proofs where simplifying expressions is key.
Even Number
Even numbers are integers divisible by 2, which means they can be written in the form \(2m\), where \(m\) is an integer. This property of being divisible by 2 means that even numbers are always in pairs, making them pivotal in equations and proofs where balance and symmetry play a role.
Recognizing even numbers is straightforward: they end in 0, 2, 4, 6, or 8. This concept is used in various mathematical contexts, including modular arithmetic, divisibility tests, and proofs. In the context of our exercise, proving an expression like \((4k^2 + 6k + 2)\) is even, involves factoring out 2, showcasing the inherent 'two-ness'.
  • Example of even numbers: 0, 2, 4, 6, ...
  • Negative even numbers: -2, -4, -6, ...
The real-world application of these numbers often involves calculations in computer science and engineering, where systems are designed based on even distributions.
Algebraic Expressions
Algebraic expressions are a cornerstone of mathematics, combining numbers and variables through operations like addition, subtraction, multiplication, and division. Expressing logical relationships, these expressions enable simplifications and transformations, which are fundamental in proving mathematical theorems.
In proofs, algebraic expressions like \(D_n = n^2 + n\) help in exploring, manipulating, and verifying mathematical properties. By considering variables like \(n\) and substituting specific values or forms such as odd integers \(2k+1\), we can simplify and derive a result, as shown in the proof in our solution. This transforms an abstract problem into a sequence of logical arithmetic operations.
  • Basic elements: variables, coefficients, constants
  • Operations: arithmetic operations forming expressions
  • Simplification: reducing expressions to simpler forms
Mastery of algebraic expressions enables the solving of equations, function analysis, and even data modeling. Their versatility is unbounded in both academic and practical scenarios.

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Most popular questions from this chapter

Using induction, prove each. $$\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin{array}{c}n+1 \\\ 1\end{array}\right)+\left(\begin{array}{c}n+2 \\\ 2\end{array}\right)+\cdots+\left(\begin{array}{c}n+r \\\ r\end{array}\right)=\left(\begin{array}{c}n+r+1 \\ r\end{array}\right) \text { (Hint: Use Pascal's identity.) }$$

Using the binomial theorem, prove each. \(\sum_{i=1}^{n}\left(\begin{array}{c}n \\\ i-1\end{array}\right)\left(\begin{array}{c}n \\\ i\end{array}\right)=\left(\begin{array}{c}2 n \\ n+1\end{array}\right)\) [Hint: Consider \((1+x)^{2 n}=(x+1)^{n}(1+x)^{n} .\) Equate the coefficients of \(\left.x^{n+1} \text { from both sides. }\right]\)

In \(1984,\) E. T. H. Wang of Wilfrid Laurier University, Waterloo, Ontario, Canada, established that $$\sum_{r=1}^{n} r^{3}\left(\begin{array}{l}n \\\r\end{array}\right) D_{n-r}=5 n !$$ Verify the formula for \(n=5\) and \(n=6\).

Using the alternate inclusion-exclusion formula, find the number of primes not exceeding: 75

The Bell numbers \(B_{n},\) named after the English mathematician Eric T. Bell (1883-1960) and used in combinatorics, are defined recursively as follows: $$B_{0}=1$$ $$B_{n}=\sum_{i=0}^{n-1}\left(\begin{array}{c} n-1 \\ i \end{array}\right) B_{i}, \quad n \geq 1$$ Compute each Bell number. $$B_{2}$$
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