Prove that ∑(1/(r^2 -1)) from r=2 to r=n is equal to (3n^2-n-2)/(4n(n+1)) for all natural numbers n>=2.

To prove the above statement we will be using proof by induction as this is the easiest way to prove that the statement holds for all natural numbers n>=2. Firstly, when proving by induction there are three main steps to consider: 1) Base case: Prove the statement holds for n = 2 2) Inductive step: Assume the statement is true for n = k and thus prove it is true for n = k+1 3) Concluding the proof 1) For n = 2: LHS: ∑(1/(r^2 -1)) = 1/(2^2-1) = 1/3 RHS: (3n^2-n-2)/(4n(n+1)) = (12-2-2)/(423) = 1/3 Thus true for n=2 2) Assume true for n=k. To prove true for n=k+1, substitute n=k+1 in LHS: ∑{r=2,k+1} (1/(r^2 -1)) = 1/((k + 1)^2 - 1) + ∑{r=2,k} (1/(r^2 - 1)) = 1/((k + 1)^2 - 1) + (3k^2 - k - 2)/(4k(k + 1)) as we have assumed true for n=k = (4(k + 1)+(3k^2 - k - 2)(k + 2))/(4k(k + 1)(k + 2)) by expanding and simplifying = (k(3k + 5))/(4(k + 1)(k + 2)) Now substitute n=k+1 in RHS: (3(k + 1)^2 - (k + 1) - 2)/(4(k + 1)((k + 1) + 1)) = (k(3k + 5))/(4(k + 1)(k + 2)) as required. Thus true for n=k+1 3) The concluding statement:From 1) the statement is true for n=2. Since the statement is true for n=k by 2) it is true for n=k+1, thus by the principle of mathematical induction it is true for all natural numbers n>=2.