Proving Mathematical Formulas Using Principle of Mathematical Induction: A Detailed Guide

Mathematical induction is a fundamental proof technique used to demonstrate the validity of statements or formulas for an infinite sequence of cases. This article delves into the application of the Principle of Mathematical Induction (PMI) to prove important summation formulas such as (sum_{k0}^n k frac{n(n 1)}{2}) and (sum_{k0}^n k^2 frac{n(n 1)(2n 1)}{6}). Understanding these proofs not only provides a rigorous foundation but also equips you to apply the same method to other complex formulas.

Key Takeaways:

Introduction to the Principle of Mathematical Induction. Proof of summation formulas using PMI. Application to proving (sum_{k0}^n k3k-1/2 n^2n1/2).

Understanding the Principle of Mathematical Induction (PMI)

The Principle of Mathematical Induction consists of two steps:

Base Case: Verify that the statement holds for the smallest value of the index, usually 0 or 1. Inductive Step: Assume the statement is true for some arbitrary value n k, and then prove that it must also be true for n k 1.

Proving the Sum of Natural Numbers

Consider the sum of the first n natural numbers:

(sum_{k0}^n k frac{n(n 1)}{2})

Step 1: Base Case

For n 1, we have:

(sum_{k0}^1 k sum_{k0}^1 k 0 1 1)

According to the formula:

( frac{1(1 1)}{2} frac{2}{2} 1)

Step 2: Inductive Step

Assume the formula is true for n k:

( sum_{k0}^k k frac{k(k 1)}{2})

Adding the next term, (k 1), to the sum:

(sum_{k0}^{k 1} k sum_{k0}^k k (k 1) frac{k(k 1)}{2} (k 1))

Factorizing the right-hand side:

(frac{k(k 1)}{2} (k 1) frac{k(k 1) 2(k 1)}{2} frac{(k 1)(k 2)}{2})

This confirms that the formula holds for n k 1.

Proving the Sum of Squares of Natural Numbers

Next, consider the sum of squares of the first n natural numbers:

(sum_{k0}^n k^2 frac{n(n 1)(2n 1)}{6})

Step 1: Base Case

For n 1, we have:

(sum_{k0}^1 k^2 0^2 1^2 1)

According to the formula:

( frac{1(1 1)(2(1) 1)}{6} frac{1(2)(3)}{6} frac{6}{6} 1)

Step 2: Inductive Step

Assume the formula is true for n k:

( sum_{k0}^k k^2 frac{k(k 1)(2k 1)}{6})

Adding the next term, (k 1)^2, to the sum:

(sum_{k0}^{k 1} k^2 sum_{k0}^k k^2 (k 1)^2 frac{k(k 1)(2k 1)}{6} (k 1)^2)

Factorizing the right-hand side:

(frac{k(k 1)(2k 1)}{6} (k 1)^2 frac{k(2k 1) 6(k 1)}{6} frac{k(2k 1) 6(k 1)}{6})

frac{k(2k 1) 6(k 1)}{6} frac{k(2k 1) 6(k 1)}{6} frac{2k^2 k 6k 6}{6} frac{(k 1)(2k 3)}{6})

This confirms that the formula holds for n k 1.

Applying Induction to the Given Summation Formula

Now, let's use induction to prove the formula:

(sum_{k0}^n k3k-1/2 n^2n1/2)

Step 1: Base Case

For n 1, we have:

(sum_{k0}^1 k3k-1/2 03(0)-1/2 13(1)-1/2 frac{0}{2} frac{1}{2} frac{1}{2})

According to the formula:

( 1^21/2 frac{1}{2})

Step 2: Inductive Step

Assume the formula is true for n k:

( sum_{k0}^k k3k-1/2 k^2k1/2)

Adding the next term, (k 1)3(k 1)-1/2, to the sum:

(sum_{k0}^{k 1} k3k-1/2 sum_{k0}^k k3k-1/2 (k 1)3(k 1)-1/2 k^2k1/2 (k 1)3(k 1)-1/2)

Simplifying the right-hand side:

(k^2k1/2 (k 1)3(k 1)-1/2 k^2k1/2 frac{(k 1)3(k 1)-1}{2})

frac{2k^2k1 (k 1)3(k 1)-1}{2} frac{2k^2k1 3(k 1)(k 1)-1}{2})

frac{2k^2k1 3(k^2 2k 1) - 1}{2} frac{2k^2k1 3k^2 6k 3 - 1}{2})

frac{5k^2 6k 2}{2} (k 1)^2(k 1)1/2)

This confirms that the formula holds for n k 1.

In conclusion, the Principle of Mathematical Induction is a powerful tool for proving complex mathematical formulas, including summation formulas. By verifying the base case and the inductive step, we can establish the validity of these formulas for all positive integers.