Question

For all k ∈N, k²+ 3k is divisible by 2

Answer 1

To prove that the expression k² + 3k is divisible by 2 for all values of k in the set of natural numbers, we can use mathematical induction.

Step-by-step explanation:

For all k ∈ N, k² + 3k is divisible by 2.

To prove that the expression k² + 3k is divisible by 2 for all values of k in the set of natural numbers, we can use mathematical induction.

1. Base case: Let's check for the smallest value of k, which is k = 1. Substituting k = 1, we get (1²) + (3 * 1) = 1 + 3 = 4, which is divisible by 2.
2. Inductive step: Assume that the expression is divisible by 2 for some arbitrary value, let's say k = n. That means (n²) + (3 * n) is divisible by 2.
3. Now, let's prove that the expression is also divisible by 2 for k = n + 1. Substituting k = n + 1, we get ((n + 1)²) + (3 * (n + 1)) = (n² + 2n + 1) + (3n + 3) = n² + 5n + 4.
4. Simplifying further, we have n² + 5n + 4 = (n² + 3n) + (2n + 4).
5. Notice that the first term (n² + 3n) is divisible by 2, as we assumed in the inductive step. The second term (2n + 4) can be rewritten as 2(n + 2), which is also divisible by 2.
6. Therefore, we have shown that both terms are divisible by 2, and hence the entire expression (n² + 5n + 4) is divisible by 2. This completes the proof by mathematical induction.

Thus, for all k ∈ N, k² + 3k is divisible by 2.