The Mathematical Induction: Mathematics Note- 2 (Essay)
Logic has deduction and induction. The former is "a method of explaining phenomena from limited principles", and the latter is "a method of discovering principles from visible phenomena". The ``mathematical induction method'' taken up here is, of course, similar to the ``induction method''. This differs from induction in that it relies heavily on the properties of the natural numbers.
In order to prove one proposition (@), the following scenario is set up.
2) If @ holds for n=r, then @ holds for n=r+1.
If these two are true, then @ is true for 1, 2, 3, 4, . . . , n and all natural numbers.
This is mathematical induction. As a test, the formula for finding (1+2+3+4+...n) is
n(n+1)/2. Let us prove this by mathematical induction.
2) If r(r+1)/2 is correct, add (r+1) to it
r(r+1)/2+(r+1)=(r+1)(r/2+1)
            =(r+1)(r+2)/2
      =(r+1)((r+1)+1)/2
The calculation was correct even when n=r+1. (Q.E.D.)
Mathematical induction has the advantage of being able to prove the proposition in question without a clear perspective. However, I think that there is an indigestion that is biased in calculation and can be proved mechanically. In addition, in the case of the example given this time, there is an eye-opening proof as a proof of "arithmetic series". up to 10
1+2+3+4+5+6+7+8+9+10 Turn it over
The sum of the corresponding terms, up and down, respectively, is (1+10), (2+9), etc. all 11.and there are 10 of them. Multiply by 110, but if you count like that, it will be a double calculation, so divide by 2,