How to proof a general expression using mathematical induction?
Answers
Mathematical Induction
Let's begin with an example.
Example: A Sum Formula
Theore. For any positive integer n, 1 + 2 + ... + n = n(n+1)/2.
Proof. (Proof by Mathematical Induction) Let's let P(n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P(n) should be an assertion that for any n is verifiably either true or false.) The proof will now proceed in two steps: the initial step and the inductive step.
Initial Step. We must verify that P(1) is True. P(1) asserts "1 = 1(2)/2", which is clearly true. So we are done with the initial step.
Inductive Step. Here we must prove the following assertion: "If there is a k such that P(k) is true, then (for this same k) P(k+1) is true." Thus, we assume there is a k such that 1 + 2 + ... + k = k (k+1)/2. (We call this the inductive assumption.) We must prove, for this same k, the formula 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2.
This is not too hard: 1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2. The first equality is a consequence of the inductive assumption.
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The Math Induction Strategy
Mathematical Induction works like this: Suppose you want to prove a theorem in the form "For all integers n greater than equal to a, P(n) is true". P(n) must be an assertion that we wish to be true for all n = a, a+1, ...; like a formula. You first verify the initial step. That is, you must verify that P(a) is true. Next comes the inductive step. Here you must prove "If there is a k, greater than or equal to a, for which P(k) is true, then for this same k, P(k+1) is true."
Since you have verified P(a), it follows from the inductive step that P(a+1) is true, and hence, P(a+2) is true, and hence P(a+3) is true, and so on. In this way the theorem has been proved.