Math Problem Statement
Use induction to show that 5^(n+1)−4n+11 is divisible by 16 for all n≥1.
Solution
We are asked to prove, by mathematical induction, that is divisible by 16 for all . This means we need to show that is divisible by 16 for every integer , i.e.,
Step 1: Base Case
We first check the base case where .
For , the expression becomes: Since is divisible by 16, the base case holds.
Step 2: Inductive Hypothesis
Now, assume that the statement holds for some arbitrary , i.e., assume that: This means:
Step 3: Inductive Step
We need to show that the statement holds for , i.e., we need to prove that: Simplifying the expression: We can group the terms as follows:
By the inductive hypothesis, we know that , so we can replace with for some integer :
Clearly, is divisible by 16, and , so:
Thus, the expression , completing the inductive step.
Conclusion:
Since both the base case and the inductive step have been proven, by mathematical induction, is divisible by 16 for all .
Would you like any more details or have further questions?
Here are five related questions that extend the idea:
- How does modular arithmetic simplify proofs in number theory?
- Can we generalize the inductive proof method for expressions with higher powers of numbers?
- How can mathematical induction be used to prove divisibility properties involving multiple variables?
- How does the inductive hypothesis relate to the structure of recursive sequences?
- What are the limitations of mathematical induction in certain types of number theory problems?
Tip: When using induction, always clearly distinguish between the base case, inductive hypothesis, and inductive step to ensure a structured approach.
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Math Problem Analysis
Mathematical Concepts
Mathematical Induction
Modular Arithmetic
Number Theory
Formulas
5^(n+1) - 4n + 11 ≡ 0 (mod 16)
Theorems
Principle of Mathematical Induction
Divisibility in Modular Arithmetic
Suitable Grade Level
Grades 11-12, College Level