Problem 28 - College Algebra

Question

In Exercises \(11-30,\) use mathematical induction to prove that each statement is true for every positive integer \(n\) 3 is a factor of \(n(n+1)(n-1)\)

Step-by-Step Solution

Answer

By mathematical induction, it is proved that for every positive integer \( n \), 3 is a factor of the product \( n(n+1)(n-1) \).

Step 1: Base case for n=1

First, check if the statement holds true for the base case of \( n = 1 \). The expression becomes \(1*(1+1)*(1-1) = 2*0 = 0\). And indeed, 3 is a factor of 0.

Step 2: Assume true for n=k

Suppose the statement is true for \( n = k \). This implies that 3 is a factor of \( k(k+1)(k-1) \). Let \( k(k+1)(k-1) = 3p \) for some positive integer \( p \). That is, \( k(k+1)(k-1) \) is divisible by 3.

Step 3: Show true for n=k+1

To complete the proof, show that the statement remains true for \( n = k+1 \). The expression becomes: \((k+1)((k+1)+1)((k+1)-1) = (k+1)(k+2)(k)\). To see if this is divisible by 3, express it in terms of \( k(k+1)(k-1) \). This can be written as \( k(k+1)(k-1) + 3k(k+1) = 3p + 3k(k+1)\), which is divisible by 3, because both terms are divisible by 3.