Introduction

Mathematical induction is a powerful technique used to prove statements about all integers or natural numbers. In this article, we will explore how to prove that the expression 10 cdot 3^n 6^{2n} is divisible by 11 for all integers n 0. We will use the principle of mathematical induction to achieve this.

Principle of Mathematical Induction

The principle of mathematical induction consists of two steps:

tBase Case: Prove that the statement holds for the smallest integer, usually n 0. tInductive Step: Assume the statement is true for some integer n k (the inductive hypothesis), and then prove that it is true for n k 1.

Proof Using Mathematical Induction

We want to prove that 10 cdot 3^n 6^{2n} is divisible by 11 for all integers n 0. Let's follow the steps of mathematical induction.

Step 1: Base Case

First, we check the base case, n 0.

Let n 0,

10 cdot 3^0 6^0 10 cdot 1 1 11, which is divisible by 11.

Step 2: Inductive Step

Assume the statement is true for n k, i.e., 10 cdot 3^k 6^{2k} is divisible by 11. We need to prove that the statement is also true for n k 1.

Consider 10 cdot 3^{k 1} 6^{2(k 1)}.

We start by rewriting the expression:

10 cdot 3^{k 1} 6^{2(k 1)} 10 cdot 3^{k 1} 36^{k 1}

10 cdot 3^{k 1} 36 cdot 6^{2k}

3 cdot 10 cdot 3^k 6^{2k} 3 cdot 11 cdot 6^{2k}

3 cdot 11 cdot m 3 cdot 11 cdot k 11 (3m 3k)

Since both 3m and 3k are integers, the expression 11 (3m 3k) is clearly divisible by 11. Therefore, by the principle of mathematical induction, the statement 10 cdot 3^n 6^{2n} is divisible by 11 for all integers n 0.

An Alternative Approach

Another way to prove this expression is to note that:

10 cdot 3^n 6^{2n} 3^n 10 12^n

Given that 3^n is never a multiple of 11, the problem reduces to showing that:

11 mid 10 12^n

When n 0, we get:

f_0 10 1 11, establishing the base case.

For the inductive step, assume for some nonnegative integer k that:

f_k 11m, m in mathbb{Z}

Then:

f_{k 1} f_k - f_{k-1} f_k 11m 10 12^{k-1} - 10 12^{k-2} 11m 11m 12^k - 12 11m 11(m 12^{k-1}

Since 11m 12^{k-1} is a multiple of 11, the statement holds for all integers n 0.

Conclusion

We have successfully proven that 10 cdot 3^n 6^{2n} is divisible by 11 for all integers n 0 using the principle of mathematical induction. This provides a rigorous mathematical proof and showcases the power of this technique in number theory.