Math Problem Statement

Use the mathematical induction principle to prove that 10^n + 3 * 4^(n+2) + 5 is divisible by 9 for all natural numbers n.

Solution

The problem is asking us to use mathematical induction to prove that the expression 10n+34n+2+510^n + 3 \cdot 4^{n+2} + 5 is divisible by 9 for all natural numbers nn.

Step 1: Base Case

For n=1n = 1:

101+341+2+5=10+343+5=10+364+5=10+192+5=20710^1 + 3 \cdot 4^{1+2} + 5 = 10 + 3 \cdot 4^3 + 5 = 10 + 3 \cdot 64 + 5 = 10 + 192 + 5 = 207

Now, check if 207 is divisible by 9:

207÷9=23207 \div 9 = 23

Since 207 is divisible by 9, the base case holds.

Step 2: Inductive Step

Assume the statement is true for n=kn = k; that is, assume:

10k+34k+2+5 is divisible by 910^k + 3 \cdot 4^{k+2} + 5 \text{ is divisible by 9}

This means:

10k+34k+2+5=9mfor some integer m.10^k + 3 \cdot 4^{k+2} + 5 = 9m \quad \text{for some integer } m.

Step 3: Prove for n=k+1n = k + 1

We need to prove that:

10k+1+34(k+1)+2+5 is divisible by 9.10^{k+1} + 3 \cdot 4^{(k+1) + 2} + 5 \text{ is divisible by 9}.

Simplify the expression:

10k+1+34k+3+510^{k+1} + 3 \cdot 4^{k+3} + 5

Rewrite 10k+110^{k+1} as 1010k10 \cdot 10^k:

1010k+34k+3+510 \cdot 10^k + 3 \cdot 4^{k+3} + 5

Substituting the inductive hypothesis (10k+34k+2+5=9m10^k + 3 \cdot 4^{k+2} + 5 = 9m):

=10(9m34k+25)+34k+3+5= 10 \cdot (9m - 3 \cdot 4^{k+2} - 5) + 3 \cdot 4^{k+3} + 5

Simplify further:

=90m304k+250+34k+3+5= 90m - 30 \cdot 4^{k+2} - 50 + 3 \cdot 4^{k+3} + 5

Notice 4k+3=44k+24^{k+3} = 4 \cdot 4^{k+2}:

=90m304k+250+124k+2+5= 90m - 30 \cdot 4^{k+2} - 50 + 12 \cdot 4^{k+2} + 5

Simplify the expression:

=90m184k+245= 90m - 18 \cdot 4^{k+2} - 45

Finally, we get:

=9(10m24k+25)= 9 (10m - 2 \cdot 4^{k+2} - 5)

Since the expression is in the form of 9×integer9 \times \text{integer}, it is divisible by 9. This completes the inductive step.

Therefore, by the principle of mathematical induction, 10n+34n+2+510^n + 3 \cdot 4^{n+2} + 5 is divisible by 9 for all natural numbers nn.

Would you like more details, or do you have any specific questions?

Here are 5 related questions to explore:

  1. How does mathematical induction work in proving divisibility problems?
  2. Can we use other methods to prove that the expression is divisible by 9?
  3. What are some common mistakes when applying mathematical induction?
  4. How does changing the base case affect the induction process?
  5. Can mathematical induction be used to prove inequalities?

Tip: When using mathematical induction, ensure your base case is correct and your inductive step logically follows the assumption.

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Divisibility
Exponents

Formulas

10^n + 3 * 4^(n+2) + 5
Divisibility by 9

Theorems

Mathematical Induction
Divisibility Rules

Suitable Grade Level

Grades 9-12