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 5n+14n+115^{n+1} - 4n + 11 is divisible by 16 for all n1n \geq 1. This means we need to show that 5n+14n+115^{n+1} - 4n + 11 is divisible by 16 for every integer n1n \geq 1, i.e.,

5n+14n+110(mod16)5^{n+1} - 4n + 11 \equiv 0 \pmod{16}

Step 1: Base Case

We first check the base case where n=1n = 1.

For n=1n = 1, the expression becomes: 51+14(1)+11=524+11=254+11=325^{1+1} - 4(1) + 11 = 5^2 - 4 + 11 = 25 - 4 + 11 = 32 Since 3232 is divisible by 16, the base case holds.

Step 2: Inductive Hypothesis

Now, assume that the statement holds for some arbitrary n=kn = k, i.e., assume that: 5k+14k+110(mod16)5^{k+1} - 4k + 11 \equiv 0 \pmod{16} This means: 5k+14k+11=16mfor some integer m.5^{k+1} - 4k + 11 = 16m \quad \text{for some integer } m.

Step 3: Inductive Step

We need to show that the statement holds for n=k+1n = k+1, i.e., we need to prove that: 5(k+1)+14(k+1)+110(mod16).5^{(k+1)+1} - 4(k+1) + 11 \equiv 0 \pmod{16}. Simplifying the expression: 5(k+1)+14(k+1)+11=5k+24(k+1)+11=55k+14k4+11.5^{(k+1)+1} - 4(k+1) + 11 = 5^{k+2} - 4(k+1) + 11 = 5 \cdot 5^{k+1} - 4k - 4 + 11. We can group the terms as follows: 55k+14k4+11=5(5k+14k+11)4.5 \cdot 5^{k+1} - 4k - 4 + 11 = 5 \cdot (5^{k+1} - 4k + 11) - 4.

By the inductive hypothesis, we know that 5k+14k+110(mod16)5^{k+1} - 4k + 11 \equiv 0 \pmod{16}, so we can replace 5k+14k+115^{k+1} - 4k + 11 with 16m16m for some integer mm: 5(16m)4=80m4.5 \cdot (16m) - 4 = 80m - 4.

Clearly, 80m80m is divisible by 16, and 412(mod16)-4 \equiv 12 \pmod{16}, so: 80m40(mod16).80m - 4 \equiv 0 \pmod{16}.

Thus, the expression 5k+24(k+1)+110(mod16)5^{k+2} - 4(k+1) + 11 \equiv 0 \pmod{16}, completing the inductive step.

Conclusion:

Since both the base case and the inductive step have been proven, by mathematical induction, 5n+14n+115^{n+1} - 4n + 11 is divisible by 16 for all n1n \geq 1.

Would you like any more details or have further questions?


Here are five related questions that extend the idea:

  1. How does modular arithmetic simplify proofs in number theory?
  2. Can we generalize the inductive proof method for expressions with higher powers of numbers?
  3. How can mathematical induction be used to prove divisibility properties involving multiple variables?
  4. How does the inductive hypothesis relate to the structure of recursive sequences?
  5. 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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