prove that the sum of the cubes of any three consecutive positive integers is di
ID: 2973123 • Letter: P
Question
prove that the sum of the cubes of any three consecutive positive integers is divisible by 3.Explanation / Answer
We are asked to show that for all n 2 Z, 3jn(n+1)(n+2). Let us use induction to establish the last assertion for all integers n 2. The base case n = 2 certainly is true. So let us take k > 2, and assume that 3j(k 1)k(k + 1). We need to show from this assumption that 3jk(k + 1)(k + 2). We compute k(k + 1)(k + 2) (k 1)k(k + 1) = k(k + 1)(k + 2 k + 1) = 3k(k + 1): Certainly 3j3k(k + 1)and hence the induction assumption that 3j(k 1)k(k + 1) guarantees that 3jk(k + 1)(k + 2) as required since the sum of two integers divisible by 3 is certainly also divisible by 3. We are left to consider the case n < 2. Notice that n(n + 1)(n + 2) = (n(n 1)(n 2)); and that for any integer a, 3ja if and only if 3 3j(a). Finally observe that n < 2 if and only if n 2 > 0 2.Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.