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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 INDUCTION.
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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 INDUCTION • One verification method in mathematics. • Commonly used to prove the theorems for all integers, especially for natural numbers.
Mathematics Induction • An important verification tools • Broadly 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”: • Basic step: prove that P(1) is true • Inductive step: Assumed that P(k) is true, it can be shown that P(k+1) is true for all k • 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)
p(1) is1 = . 1. (2), which means1 = 1, completely true • It is assumed that p(k) is true for one natural number k, which is 1+2+3+… +k = k(k+1) true • Next, it must be proved that p(k+1) is true, which is: 1+2+3+… +k + (k+1) = (k+1) (k+2)
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
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 true • Step (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:
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:
