Mathematical induction
It is normally used to solve a statement relaying upon a natural number n.
Type I: If P(n) is a statement relaying upon n, then to solve it by induction, we continue as follows:
(i) Validate the validity of P(n) for n = 1.
(ii) Suppose that P(n) is true for some positive integer m and then using it start the validity of P(n) for n = m + 1.
Then, P(n) is true for each and every n∈N.
Type II: If P(n) is a statement relaying upon n but beginning with some positive integer k, then to solve P(n), we continue as follows:
(i) Check the validity of P(n) for n = k.
(ii) Suppose that the statement is true for n = m ≥ k. Then, using it start the checking of P(n) for n = m + 1.
Then, P(n) is true for each n ≥ k
Illustration: Solve that if sin a ¹ 0, in that case
, holds for each n∈n.
Solution: If P(n) shows the provided statement, then for n = 1, P(1):
, which is true
because
![863_Mathematical induction3.png](https://www.expertsmind.com/../CMSImages/863_Mathematical induction3.png)
= cosα cos2α.
Suppose that P(n) is true for any positive integer m,
i.e. ![648_Mathematical induction4.png](https://www.expertsmind.com/../CMSImages/648_Mathematical induction4.png)
Using (1), we will solve P(n) is true for n = m + 1
![17_Mathematical induction5.png](https://www.expertsmind.com/CMSImages/17_Mathematical%20induction5.png)
Therefore, P(n) is true for each n.
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