Prove by induction that f(n)=2^(n+2)+3^(2n+1) is divisible by 7 for all positive integers n

To prove that \( f(n) = 2^{n+2} + 3^{2n+1} \) is divisible by 7 for all positive integers \( n \), we will use **mathematical induction**.

### Step 1: Base Case
We need to check if the statement holds for \( n = 1 \).

For \( n = 1 \),
\[
f(1) = 2^{1+2} + 3^{2(1) + 1} = 2^3 + 3^3 = 8 + 27 = 35.
\]
Clearly, \( 35 \) is divisible by 7, since \( 35 \div 7 = 5 \).

Thus, the base case is true.

### Step 2: Inductive Hypothesis
Assume that the statement is true for some arbitrary positive integer \( k \), i.e., assume:
\[
f(k) = 2^{k+2} + 3^{2k+1}
\]
is divisible by 7. This means:
\[
2^{k+2} + 3^{2k+1} \equiv 0 \pmod{7}.
\]

### Step 3: Inductive Step
We need to prove that the statement holds for \( k+1 \), i.e., we need to show:
\[
f(k+1) = 2^{(k+1)+2} + 3^{2(k+1)+1} = 2^{k+3} + 3^{2k+3}
\]
is divisible by 7.

We begin by expressing \( f(k+1) \):
\[
f(k+1) = 2^{k+3} + 3^{2k+3}.
\]
We can rewrite this as:
\[
f(k+1) = 2^{k+2} \cdot 2 + 3^{2k+1} \cdot 3^2.
\]
From our inductive hypothesis, we know that:
\[
2^{k+2} + 3^{2k+1} \equiv 0 \pmod{7}.
\]
Now, we need to consider \( f(k+1) \mod 7 \) in more detail.

First, note that:
\[
2^3 = 8 \equiv 1 \pmod{7} \quad \text{and} \quad 3^2 = 9 \equiv 2 \pmod{7}.
\]
Thus, the expression for \( f(k+1) \) modulo 7 becomes:
\[
f(k+1) = 2 \cdot (2^{k+2}) + 2 \cdot (3^{2k+1}).
\]
Since \( 2^{k+2} + 3^{2k+1}\equiv 0\

Explore a detailed proof by induction showing that the function f(n) = 2^(n+2) + 3^(2n+1) is divisible by 7 for all positive integers. Learn key concepts such as mathematical induction, modular arithmetic, and exponentiation properties. Suitable for advanced high school students (Grades 11-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation