Mathematical Induction PART 2

In Part 1

It shows that

Step 1 is Show it is true for n=1

Step 2 is Show that if n=k is true then n=k+1 is also true

so How to Do it?

Step 1 : * prove* it is true for n=1 (ｎｏｒｍａｌｌｙ）

Step 2 　：　 done　in this way normally we can prove it out..

First ~ Assume it is true forn=k

Second ~ Prove it is true for n=k+1 (normally we can use the n=k case as fact )

S (n / k) must be all positive integers including n/k

EXAMPLE

PROVE 1 + 3 + 5 + ... + (2n-1) = n^2

Fisrt : Show it is true for n=1

from LEFT : 1+3+5+.....+(2(1)-1) = 1

from RIGHT n^2 = (1^2) =1 SO 1 = 1^2 is True

SECOND : Assume it is true for n=k

1 + 3 + 5 + ... + (2^k-1) = k^2 is True

(prove it by yourself :) write down your steps in comment )

THIRD : prove it is true for k+1

1 + 3 + 5 + ... + (2^k-1) + (2^(k+1)-1) = (k+1)^2 ... ? (prove it by yourself :) write down your steps in comment )

We know that 1 + 3 + 5 + ... + (2k-1) = k^2 (the assumption above), so we can do a replacement for all but the last term:

k^2 + (2(k+1)-1) = (k+1)^2

THEN expanding all terms:

LEFT : k^2 + 2k + 2 - 1 = k^2 + 2k+1

NEXT simplifying :

RIGHT :k^2 + 2k + 1 = k^2 + 2k + 1

LEFT and RIGHT are the same! So it is true, it is proven.

THEREFORE :

1 + 3 + 5 + ... + (2(k+1)-1) = (k+1)^2 is TRUE!!!!!!

Mathematical Induction IS done !!! :)

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