Prove that n3-n is divisible by 6 for all n 1.

Here is the proof using the Principle of Mathematical Induction. Prove that n^3-n is divisible by 6 for all n 1. Let P(n) be the statement "n^3-n is divisible by 6". Base Case (n=1): P(1): 1^3-1 = 1-1 = 0. Since 0 is divisible by 6 (0 = 6 × 0), P(1) is true. Inductive Hypothesis: Assume P(m) is true for some positive integer m. That is, assume m^3-m is divisible by 6. This means m^3-m = 6k for some integer k. Inductive Step: We need to prove that P(m+1) is true, i.e., (m+1)^3-(m+1) is divisible by 6. Consider the expression (m+1)^3-(m+1): (m+1)^3-(m+1) Expand the term (m+1)^3: (m^3+3m^2+3m+1) - (m+1) Distribute the negative sign: m^3+3m^2+3m+1 - m - 1 Combine like terms: m^3+3m^2+2m Rearrange the terms to use the inductive hypothesis: (m^3-m) + 3m^2+3m By the Inductive Hypothesis, m^3-m = 6k. Substitute this into the expression: 6k + 3m^2+3m Factor out 3m from the last two terms: 6k + 3m(m+1) We know that 6k is divisible by 6. Now consider the term 3m(m+1). The product of two consecutive integers, m(m+1), is always an even number (because either m is even or m+1 is even). Therefore, m(m+1) can be written as 2j for some integer j. Substitute this into the term: 3(2j) = 6j So, 3m(m+1) is divisible by 6. Since both 6k and 3m(m+1) are divisible by 6, their sum is also divisible by 6: 6k + 6j = 6(k+j) This shows that (m+1)^3-(m+1) is divisible by 6. Thus, P(m+1) is true. By the Principle of Mathematical Induction, P(n) is true for all n 1. The proof is complete.