Answer:
We will prove the statement by mathematical induction.
Base case: For n=2, we have 5(2) + 9 = 19 and 6(2) = 12. Clearly, 19 ≤ 12 is false, so the base case does not satisfy the inequality.
Inductive step: Assume that for some integer k ≥ 2, 5k + 9 ≤ 6k holds. We will show that 5(k+1) + 9 ≤ 6(k+1) also holds.
Starting with the left-hand side of the inequality:
5(k+1) + 9 = 5k + 5 + 9 = 5k + 14
Since k ≥ 2, we can use the inductive hypothesis to write:
5k + 9 ≤ 6k
Adding 5 to both sides:
5k + 14 ≤ 6k + 5
Substituting back into the left-hand side of the original inequality:
5(k+1) + 9 ≤ 6k + 5 = 6(k+1)
Therefore, we have shown that if the statement holds for some integer k ≥ 2, then it also holds for k+1.
Conclusion: By the principle of mathematical induction, the statement 5n + 9 ≤ 6n holds for all integers n ≥ 2.
rate5* po and give thanks for more! your welcome po!