Final answer:
To prove the equation, we can use a mathematical induction proof. Plug in the base case of n=1 to show the equation holds true for n=1. Assume the equation holds true for n, and substitute n+1 for n to show it holds true for n+1. Simplify both sides to show they are equal. Therefore, the equation holds true for all positive integers n.
Step-by-step explanation:
To prove that for every positive integer n, the equation 1 · 2 · 3 ... (n(n-1)(n-2) = n(n-1)(n-2)(n-3)/4 holds true, we can use a mathematical induction proof. Mathematical induction is a technique used to prove a statement is true for all positive integers by showing it holds true for a base case, and then showing that if it holds true for n, it also holds true for n+1. In this case, the base case would be n=1.
To start the proof, we can plug in n=1 into both sides of the equation:
Left side: 1(1-1)(1-2) = 1(0)(-1) = 0
Right side: 1(1-1)(1-2)(1-3)/4 = 1(0)(-1)(-2)/4 = 0
Since both sides of the equation are equal when n=1, we have shown the base case is true. Now, we need to show that if it holds true for n, it also holds true for n+1. We can assume the equation holds true for n, so:
1 · 2 · 3 ... (n(n-1)(n-2) = n(n-1)(n-2)(n-3)/4
Next, we want to show that it holds true for n+1. We can substitute n+1 in place of n:
1 · 2 · 3 ... ((n+1)n(n-1)(n-2) = (n+1)n(n-1)(n-2)(n-3)/4
Now, we can simplify both sides of the equation:
Left side: ((n+1)n(n-1)(n-2) = (n+1)n(n-1)(n-2)(n-3)
Right side: (n(n-1)(n-2)(n-3)/4) * (n+1) = (n(n-1)(n-2)(n-3)(n+1))/4
Since both sides of the equation simplify to the same expression, we have shown that if the equation holds true for n, it also holds true for n+1. Therefore, the equation holds true for all positive integers n by using mathematical induction.