You can still substitute k(k+1) using k(k-1)(k+1)=2F.sooyong94 said:2F+3k(k+1)
Yes. Now you should see why it is an even integer.sooyong94 said:2F+3(2F/(k-1) ?
I don't know if that's a typo or on purpose, but the LHS and RHS cannot be equal. Moreover, apart from being divisible by two you should also be convinced that 2F+6F/(k-1) is indeed an integer.sooyong94 said:2F= 2F+6F/(k-1),
Induction is a mathematical proof technique used to prove a statement for all natural numbers. In this case, we will use induction to prove that the expression r(r-1)(r+1) is always an even integer for all values of r.
The first step is to prove that the statement is true for the base case, which is usually r = 1. In this case, we can see that 1(1-1)(1+1) = 0, which is an even integer. So, the statement holds true for the base case.
The induction hypothesis is the assumption that the statement holds true for some arbitrary value of r, let's call it k. So, we assume that k(k-1)(k+1) is an even integer. We will use this assumption to prove that the statement also holds true for k+1.
To prove the statement for k+1, we substitute k+1 in place of r in the expression and simplify it. So, we have (k+1)((k+1)-1)((k+1)+1) = (k+1)(k)(k+2) = k(k+1)(k+2). Using the induction hypothesis, we know that k(k-1)(k+1) is an even integer. And since 2 is a factor of k(k+1)(k+2), it is also an even integer. Thus, the statement holds true for k+1.
No, we can only use induction to prove a statement for natural numbers. This is because the induction hypothesis relies on the property that the statement holds true for a previous value of r. Since real numbers are not discrete, we cannot use induction to prove this statement for all real numbers.