Proving a Formula for Consecutive Products Using Mathematical Induction

In this article, we will delve into the process of proving a specific formula using mathematical induction. The formula we will focus on is:

1middot;2middot;3 2middot;3middot;4 3middot;4middot;5 ... n(n 1)(n 2) frac{n(n 1)(n 2)(n 3)}{4}

Introduction to Mathematical Induction

Mathematical induction is a powerful method of mathematical proof used to establish the truth or validity of statements over an infinite sequence of cases. It is particularly useful when dealing with sums, products, and recursive definitions. This article will demonstrate an induction proof for the formula involving consecutive products of natural numbers.

Formula and Preliminary Knowledge

To prove the formula, we need to use the following preliminary knowledge:

sum_{n1}^{k} n frac{k(k 1)}{2} sum_{n1}^{k} n^2 frac{k(k 1)(2k 1)}{6}

Proof by Mathematical Induction

We will use mathematical induction to prove the given formula for all natural numbers n in mathbb{N}.

Base Case

First, we need to show that the formula is true for n1.

Base case: n  1LHS  1 middot; 2 middot; 3  6RHS  frac{1 middot; 2 middot; 3 middot; 4}{4}  frac{24}{4}  6

The formula holds for n1.

Inductive Step

Next, we assume that the formula is true for some arbitrary natural number nk. That is:

1 middot; 2 middot; 3 2 middot; 3 middot; 4 ... k(k 1)(k 2) frac{k(k 1)(k 2)(k 3)}{4}

Now, we need to prove that the formula is also true for nk 1. We start with the left-hand side for nk 1 and use the inductive hypothesis:

Step 1: Expand the formula for nk 1

1 middot; 2 middot; 3 2 middot; 3 middot; 4 ... k(k 1)(k 2) (k 1)(k 2)(k 3)

Using the inductive hypothesis, we get:

1 middot; 2 middot; 3   2 middot; 3 middot; 4   ...   k(k 1)(k 2)   (k 1)(k 2)(k 3) frac{k(k 1)(k 2)(k 3)}{4}   (k 1)(k 2)(k 3)

Step 2: Combine the terms on the right-hand side

Factor out the common term (k 1)(k 2)(k 3) from the right-hand side:

frac{k(k 1)(k 2)(k 3)}{4}   (k 1)(k 2)(k 3) (k 1)(k 2)(k 3) left( frac{k}{4}   1 right)

Simplify the expression inside the parentheses:

(k 1)(k 2)(k 3) left( frac{k   4}{4} right) frac{(k 1)(k 2)(k 3)(k 4)}{4}

This is exactly the right-hand side of the formula for nk 1. Therefore, the formula is true for nk 1.

Conclusion

By mathematical induction, we have proven that the formula is true for all natural numbers n in mathbb{N}:

1 middot; 2 middot; 3 2 middot; 3 middot; 4 ... n(n 1)(n 2) frac{n(n 1)(n 2)(n 3)}{4}

References

For more information on mathematical induction and its applications in various mathematical proofs, refer to the following resources:

Wikipedia: Mathematical Induction MathWorld: Mathematical Induction

This article provides a step-by-step guide to proving the given formula using mathematical induction. Understanding these concepts is crucial for solving various mathematical problems and for those pursuing advanced studies in mathematics.