Can you explain the approach to solving IB maths induction questions?

Proof by maths induction is a type of proof used in mathematical analysis to demonstrate that a certain expression or statement P(n) involving an arbitrary neutral number n is true for all n belonging to N. The most commonly used approach to solving IB maths induction questions is splitting the proof into 4 parts as follows:1. Check that a statement P(n) is true, for example, in the case when n=1 (it does not necessarily have to equal to 1. It depends on the statement, the goal here is to use the first natural number n for which the given expression holds). 2. Assume that P(n) is true for some n=k. Basically, we are assuming here that the statement is true for some arbitrary number k.3. Now we show that it is also true for n=k+1. This is the most complicated step of the 4 - the actual proof. There are many methods that one could use, all of which depend on the expression given. However, in most cases you would want to use some information, obtained from the previous step, where you assumed P(n) to be true for n=k, in order to get to the same expression as for n=k+1. 4. We must know formally conclude that if the statement P(n) is true for n=1 and is also true for n=k+1, assuming true for n=k, then by the principles of mathematical induction the statement P(n) must be true for all n belonging to N.  Let us consider an example. Prove that E(k) = (1/2)n(n+1), where E is the sum of all natural numbers from k=1 to n. 1. Let us check if the statement is true for n=1: P(1)= 1 = (1/2)1(2) = 1. So P(n) is indeed true for the case when n=1.2. Now let is assume that P(n) is true for n=k: then P(k)= (1/2)k(k+1).3. Let us show that P(n) is also true for n=k+1: P(k+1) = E(k+1) = (1/2)(k+1)(k+2). Using the assumption that we made in the previous step we see that the sum of the first k+1 terms can be rewritten as the sum of the first k terms plus the (k+1) term, i.e. E(k=1) = E(k) + (k+1), so (1/2)k(k+1) + (k+1) = (k+1)(1/2k+1) = (1/2)(k+1)(k+2) = E(k+1).4. Therefore, since it is true for n=1, also true for n=k+1 assuming true for n=k, by the principles of mathematical induction the statement E(k) = (1/2)n(n+1) is true for all n that belong to N.