To prove that n^3 + 2n is divisible by 3 using mathematical induction, we must first show that the formula holds true for the base case, n = 1.
When n = 1, we have:
1^3 + 2(1) = 1 + 2 = 3
Since 3 is divisible by 3, the base case is true.
Next, we assume that the formula is true for some arbitrary value of n, and we must show that it holds true for n+1.
Assuming that n^3 + 2n is divisible by 3, we consider:
(n+1)^3 + 2(n+1)
Expanding the cube, we get:
n^3 + 3n^2 + 3n + 1 + 2n + 2
Simplifying, we get:
(n^3 + 2n) + (3n^2 + 3n + 3)
Since we know that n^3 + 2n is divisible by 3 (by our assumption), we only need to show that 3n^2 + 3n + 3 is also divisible by 3.
Factoring out a 3 from the expression, we get:
3(n^2 + n + 1)
Since n^2 + n + 1 is always an integer, the entire expression is divisible by 3.
Therefore, we have shown that if the formula is true for n, it must also be true for n+1, and since the formula is true for n=1 (our base case), the formula is true for all positive integers n.
Hence, we have proven that n^3 + 2n is divisible by 3 using mathematical induction.