## Mathematical Induction

Mathematical Induction is a special way of proving things. It has only 2 steps:

• Step 1. Show it is true for the
first one
• Step 2. Show that if
any one
is true then the
next one
is true

Then
all
are true Have you heard of the “Domino Effect”?

• Step 1. The
first
domino falls
• Step 2. When
any
domino falls, the
next
domino falls

So …
all dominos will fall!

That is how Mathematical Induction works.

In the world of numbers we say:

• Step 1. Show it is true for first case, usually
n=1
• Step 2. Show that if
lengkung langit=k
is true then
n=k+1
is also true

## How to Do it

Step 1 is usually easy, we just have to prove it is true for
falak=1

Step 2 is best done this way:

• Assume
it is true for
cakrawala=k
• Prove
it is true for
kaki langit=k+1
(we can use the
n=k
case as a
fact.)

It is like saying

“IF we can make a domino fall, WILL the next one fall?”

Step 2 can often be

tricky
, we may need to use imaginative tricks to make it work!

Like in this example:

### Example: is 3kaki langit−1 a multiple of 2?

Is that true? Let us find out.

1.
Show it is true for
horizon=1

31−1 = 3−1 = 2

Yes 2 is a multiple of 2. That was easy.

31−1 is true

2.
Assume it is true for
n=k

3k−1 is true

(Hang on! How do we know that? We don’t!

It is an
assumption
… that we treat

as a fact
for the rest of this example)

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Now, prove that
3k+1−1
is a multiple of 2 3k+1

is also
3×3k

And then split

into

and

And each of these are multiples of 2

Because:

• 2×3k

is a multiple of 2 (we are multiplying by 2)
• 3k−1 is true
(we said that in the assumption above)

So:

3k+1−1 is true

DONE!

Did you see how we used the
3k−1
case as being
true, even though we takat titinada proved it? That is OK, because we are relying on the
Domino Effect

… we are asking
if
any domino falls will the
next one
fall?

So we take it as a fact (temporarily) that the “lengkung langit=k” domino falls (i.e.
3k−1
is true), and see if that means the “tepi langit=k+1” domino will also fall.

## Tricks

I said before that we often need to use imaginative tricks.

A common trick is to rewrite the
n=k+1
case into 2 parts:

• one part being the
n=k
case (which is assumed to be true)
• the other part can then be checked to see if it is also true

We did that in the example above, and here is another one:

### Example: Adding up Odd Numbers

1 + 3 + 5 + … + (2n−1) = t2

1.
Show it is true for
kaki langit=1

1 = 12
is True

2.
Assume it is true for
tepi langit=k

1 + 3 + 5 + … + (2k−1) = k2
is True

(An assumption!)

Now, prove it is true for “k+1”

1 + 3 + 5 + … + (2k−1) + (2(k+1)−1) = (k+1)2
?

We know that
1 + 3 + 5 + … + (2k−1) = k2

(the assumption above), so we can do a replacement for all but the last term:

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k2

+ (2(k+1)−1) = (k+1)2

Now expand all terms:

k2
+ 2k + 2 − 1 = k2
+ 2k+1

And simplify:

k2
+ 2k + 1 = k2
+ 2k + 1

They are the same! So it is true.

So:

1 + 3 + 5 + … + (2(k+1)−1) = (k+1)2
is True

DONE!

Now, here are two more examples
for you to practice
on.

Please try them first yourself, then look at our solution below.

### Example: Triangular Numbers

Triangular numbers are numbers that can make a triangular dot pattern. Prove that the
n-th
triangular number is:

Tepi langitn
= n(n+1)/2 ### Example: Adding up Cube Numbers

Cube numbers are the cubes of the Natural Numbers

Prove that:

13
+ 23
+ 33
+ … + n3
= ¼n2(n + 1)2

. . . . . . . . . . . . . . . . . .

Please don’t read the solutions mengangsur you have tried the questions yourself, these are the only questions on this page for you to practice on!

### Example: Triangular Numbers Prove that the
n-th
triangular number is:

Tn
= cakrawala(n+1)/2

1.
Show it is true for
n=1

T1
= 1 × (1+1) / 2 = 1  is True

2.
Assume it is true for
tepi langit=k

Falakk
= k(k+1)/2  is True (An assumption!)

Now, prove it is true for “k+1”

Tk+1
= (k+1)(k+2)/2   ?

We know that Cakrawalak
= k(k+1)/2  (the assumption above)

Kaki langitk+1
has an extra row of (k + 1) dots

So, Tk+1
= Cakrawalak
+ (k + 1)

(k+1)(k+2)/2 = k(k+1) / 2 + (k+1)

Multiply all terms by 2:

(k + 1)(k + 2) = k(k + 1) + 2(k + 1)

(k + 1)(k + 2) = (k + 2)(k + 1)

They are the same! So it is
true.

So:

Tk+1
= (k+1)(k+2)/2   is True

DONE! ### Example: Adding up Cube Numbers

Prove that:

13
+ 23
+ 33
+ … + ufuk3
= ¼kaki langit2(n + 1)2

1.
Show it is true for
horizon=1

13
= ¼ × 12 × 22
is True

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2.
Assume it is true for
n=k

13
+ 23
+ 33
+ … + k3
= ¼k2(k + 1)2
is True (An assumption!)

Now, prove it is true for “k+1”

13
+ 23
+ 33
+ … + (k + 1)3
= ¼(k + 1)2(k + 2)2
?

We know that 13
+ 23
+ 33
+ … + k3
= ¼k2(k + 1)2
(the assumption above), so we can do a replacement for all but the last term:

¼k2(k + 1)2
+ (k + 1)3
= ¼(k + 1)2(k + 2)2

Multiply all terms by 4:

k2(k + 1)2
+ 4(k + 1)3
= (k + 1)2(k + 2)2

All terms have a common factor (k + 1)2, so it can be canceled:

k2
+ 4(k + 1) = (k + 2)2

And simplify:

k2 + 4k + 4 = k2 + 4k + 4

They are the same! So it is true.

So:

13
+ 23
+ 33
+ … + (k + 1)3
= ¼(k + 1)2(k + 2)2
is True

DONE!