Problem 11
Question
Prove each. \(D_{n}\) is even if \(n\) is an odd integer.
Step-by-Step Solution
Verified Answer
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.
1Step 1: 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.
2Step 2: 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 \(D_n\).
\[D_{2k+1} = (2k+1)^{2}+(2k+1)\]
3Step 3: 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}
4Step 4: 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.
5Step 5: Conclusion
Therefore, we have shown that for any odd integer value of \(n\), \(D_n\) will be even, as proven with the expression \(2(2k^{2}+3k+1)\).
Key Concepts
Odd IntegerEven NumberAlgebraic Expressions
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.
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.
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'.
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, ...
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.
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
Other exercises in this chapter
Problem 11
Find the number of solutions to each equation with non-negative integer variables. \(x+y+z=11, x \leq 3, y \leq 4, z \leq 5\)
View solution Problem 11
Find the number of ternary words over the alphabet \\{0,1,2\\} that are of length four and: Contain at most two \(0^{\prime} s.\)
View solution Problem 11
Let \(A\) be a 10 -element subset of the set \(\\{1,2, \ldots, 20\\}\) Determine if \(A\) has two five-element subsets that yield the same sum of the elements.
View solution Problem 11
Find the number of palindromic alphanumeric identifiers of length \(n\).
View solution