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 for n=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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## Comments

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TopNewestcalculus?

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so it is math

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really?

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what

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roar

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feeling lucky

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feeling happy

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feeling sad

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tiger

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is that true?

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yes

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no

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What have you been telling?

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what is your question ? I can't understand what did you want to say through your comments..

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a

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good

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nice

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