If n is a positive integer, then `n^(3)+2n` is divisible by

Options

A. 2
B. 6
C. 15
D. 3

Answer

To determine if \( n^3 + 2n \) is divisible by 3 for any positive integer \( n \), we can follow a structured approach using mathematical induction.

### Step-by-Step Solution:

1. **Base Case:**
We start by checking the base case when \( n = 1 \).
\[ f(1) = 1^3 + 2 \times 1 = 1 + 2 = 3 \]
Since 3 is divisible by 3, the base case holds.

2. **Inductive Hypothesis:**
Assume that for some positive integer \( k \), \( f(k) = k^3 + 2k \) is divisible by 3. This means there exists an integer \( m \) such that:
\[ f(k) = k^3 + 2k = 3m \]

3. **Inductive Step:**
We need to show that \( f(k + 1) \) is also divisible by 3. We compute:
\[ f(k + 1) = (k + 1)^3 + 2(k + 1) \]
Expanding this expression:
\[ f(k + 1) = (k^3 + 3k^2 + 3k + 1) + 2k + 2 \]
Simplifying it:
\[ f(k + 1) = k^3 + 3k^2 + 5k + 3 \]

4. **Rearranging:**
We can rearrange \( f(k + 1) \) as follows:
\[ f(k + 1) = (k^3 + 2k) + 3(k^2 + k + 1) \]
From our inductive hypothesis, we know that \( k^3 + 2k \) is divisible by 3 (i.e., \( f(k) = 3m \)). Therefore, we can write:
\[ f(k + 1) = 3m + 3(k^2 + k + 1) \]
Factoring out the 3:
\[ f(k + 1) = 3(m + k^2 + k + 1) \]
This shows that \( f(k + 1) \) is also divisible by 3.

5. **Conclusion:**
By the principle of mathematical induction, since the base case holds and the inductive step has been proven, we conclude that \( n^3 + 2n \) is divisible by 3 for all positive integers \( n \).