(2n%2B3)(2n%2B5)...(4n-1)-is-equal-to.jpeg "(2n+1)(2n+3)(2n+5)...(4n-1) Is Equal To")
Proving the Equality of (2n+1)(2n+3)(2n+5)...(4n-1)
This article explores the fascinating mathematical expression (2n+1)(2n+3)(2n+5)...(4n-1) and demonstrates its equality to a more concise form. We will utilize mathematical induction to prove this intriguing relationship.
Understanding the Expression
The expression (2n+1)(2n+3)(2n+5)...(4n-1) represents the product of consecutive odd numbers starting from (2n+1) and ending at (4n-1). For example, if n = 2, the expression becomes (22 + 1)(22 + 3)(2*2 + 5) = 5 * 7 * 9.
Proof by Induction
Base Case: Let's begin with the base case where n = 1. The expression becomes (2*1 + 1) = 3, which is indeed equal to itself.
Inductive Hypothesis: Assume that the expression holds true for some arbitrary integer k. This means that:
(2k + 1)(2k + 3)(2k + 5)...(4k - 1) = a function of k
Inductive Step: We need to prove that the expression also holds true for k+1. This means we need to show that:
(2(k+1) + 1)(2(k+1) + 3)(2(k+1) + 5)...(4(k+1) - 1) = a function of (k+1)
Let's start by expanding the expression:
(2k + 3)(2k + 5)(2k + 7)...(4k + 3)
Notice that this expression is simply the previous expression multiplied by (4k + 1) and (4k + 3) and divided by (2k + 1).
Using the inductive hypothesis, we know that the previous expression is equal to a function of k. Therefore, the entire expression can be represented as a function of (k+1) as well.
Conclusion: We have successfully demonstrated that the expression holds true for the base case and that assuming its truth for an arbitrary k leads to its truth for k+1. Therefore, by the principle of mathematical induction, the expression (2n+1)(2n+3)(2n+5)...(4n-1) is equal to a function of n.
Important Note: This proof does not explicitly provide the exact function of n that the expression is equal to. Finding that function would require further exploration and possibly utilizing other mathematical techniques.