MATHEMATICS INDUCTION AND BINOM THEOREM

By : IRA KURNIAWATI, S.Si, M.Pd

Competence Standard

Able to:Understand and prove the

theorem using Mathematics Induction

Apply Binom theorem in the descriptions of the form of power (a+b)n

MATHEMATICS INDUCTIONOne verification method in

mathematics.Commonly used to prove the

theorems for all integers, especially for natural numbers.

Mathematics InductionAn important verification toolsBroadly used to prove

statements connected with discreet objects (algorithm complexity, graph theorems, identity and inequality involving integers, etc.)

Cannot be used to find/invent theorems/formula, and can only be used to prove something

Mathematics Induction:A technique to prove proposition in the form of n P(n), in which the whole discussion is about positive integers sets

Three steps to prove (using mathematics induction) that “P(n) is true for all n positive integers”:

1. Basic step: prove that P(1) is true2. Inductive step: Assumed that P(k)

is true, it can be shown that P(k+1) is true for all k

3. Conclusion: n P(n) is true

The Steps to prove the theorems using mathematics induction are :Supposing p(n) is a statement

that will be proved as true for all natural numbers.

Step (1) : it is shown that p(1) is true.

Step (2) : it is assumed that p(k) is true for k natural number and it is shown that p(k+1) is true.

When steps 1 and 2 have been done correctly, it can be concluded that p(n) is correct for all n natural number

Step (1) is commonly called as the basic for induction

Step (2) is defined as inductive step.

Example:Using Mathematics Induction,

prove that 1+2+3+…+n= n(n+1) for all n natural number

Prove:Suppose p(n) declares1+2+3+…

+n= n(n+1)

2

1

2

1

(i) p(1) is1 = . 1. (2), which means1 = 1, completely true

(ii) It is assumed that p(k) is true for one natural number k, which is 1+2+3+… +k = k(k+1) true

(iii)Next, it must be proved that p(k+1) is true, which is:

1+2+3+… +k + (k+1) = (k+1) (k+2)

2

1

2

1

2

1

It should be shown as follows:1+2+3+… +k + (k+1) = (1+2+3+…+k)

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

So:1+2+3+… +k + (k+1) = (k+1) (k+2)

which means that p(k+1) is true.

It follows that p(n) is true for all n natural number

2

1

2

1

2

1

2

1

Example 2:

Show that n < 2n for all positive n natural number.

Solution:Suppose P(n): proposition “n < 2n”

Basic step: P(1) is true, because 1 < 21 = 2

Inductive step: Assumed that P(k) is true for all k natural number, namely k < 2k

We need to prove that P(k+1) is true, which is: k + 1 < 2k+1

Start from k < 2k

k + 1 < 2k + 1 2k + 2k = 2k+1

So, if k < 2k, then k + 1 < 2k+1 P(k+1) is true

Conclusion:n < 2n is true for all positive n natural

number

The basic of induction is not always taken from n=1; it can be taken as suited to the problems encountered or to statements to be proved

Supposing p(n) is true for all natural numbers n ≥ t.

The steps to prove it using mathematics induction are:

Step (1) : show that p(t) is trueStep (2) : assume that p(k) is

true for natural number k ≥ t, and show that p(k+1) is true

Binom Theorem

The combination of r object taken from n object, exchanged with C(n,r) or and formulated as:

r

n

)!(!

!

rnr

n

r

n

Example:

Suppose there are 5 objects, namely a,b,c,d, and e. If out of these 5 objects 3 are taken away, the ways to take those 3 objects are:

ways

r

n10

)1.2.3)(1.2(

1.2.3.4.5

!3!2

!5

The property of Binom Coefisient

n

n

n

nnnnso

n

nnnni

2...210

...210

11

)(

!)!(

!

)!(!

!)(

sticcharacterilsymmetricakn

n

k

nso

kkn

n

kn

nand

knk

n

k

nii

1

11

thenk, n and numbers natural arek andn If (iii)

k

n

k

n

k

n

mk

mn

m

n

m

k

k

n

so k,mn and numbers natural arek and m, n, If (iv)

1

1

so k, n and numbers natural arek andn If (v)

k

nn

k

nk

1

1...

2

2

1

1

0

1

1...

21

r

rk

r

rkkkk

k

n

k

n

k

k

k

k

k

kvi

n

n

n

nnnnvii

2...

210)(

2222

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