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       

            = cosα cos2α.

Suppose that P(n) is true for any positive integer m,

i.e.       

Using (1), we will solve P(n) is true for n = m + 1

Therefore, P(n) is true for each n.

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