Mathematical induction
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Transcript of Mathematical induction
PRINCIPLE OF MATHEMATICAL
INDUCTION
WHAT IS
MATHEMATICAL
INDUCTION?
Mathematical induction is one of the techniques which can be
used to prove varietyof mathematical statements
which are formulated in terms of n, where n is apositive integer.
HOW TO PROVE IT?
Suppose there is a given statement P(n) involving the natural number n such that:-
1. The statement is true for n = 1, i.e., P(1) is true, and2. If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P(k) implies the truth of P (k + 1).
Then, P(n) is true for all natural
numbers n.
Property (i) is simply a statement of fact. There may
be situations when a statement is true for all n ≥ 4. In this case, step 1 will start
from n = 4 and we shall verify the result for n = 4, i.e., P(4).
Property (ii) is a conditional property. It does not assert that
the given statement is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1. So, to prove that the property holds ,
only prove that conditional proposition:
If the statement is true for n = k, then it is also true for n = k + 1.This is sometimes referred to as
the inductive step. The assumption that the given statement is true for n = k in this inductive step is called
the inductive hypothesis.
Let us take a look at Some examples:-
Let n = 1. Then:2 + 22 + 23 + 24 + ... + 2n = 21 = 2...and:2n+1 – 2 = 21+1 – 2 = 22 – 2 = 4 – 2 = 2So P(n) works for n = 1.
Qs 1. For n > 1, prove:-P(n)=2 + 22 + 23 + 24 + ... + 2n = 2n+1 – 2
Assume, for n = k, that P(n) holds; that is, that 2 + 22 + 23 + 24 + ... + 2k = 2k+1 – 2Let n = k + 1.2 + 22 + 23 + 24 + ... + 2k + 2k+1 = [2 + 22 + 23 + 24 + ... + 2k] + 2k+1 = [2k+1 – 2] + 2k+1 = 2×2k+1 – 2 = 21×2k+1 – 2 = 2k+1+1 – 2 = 2(k+1)+1 – 2Then P(n) works for n = k + 1.Thus P(n) is true for n ≥ 1
Qs. 2 For all n ≥ 1, prove that12 + 22 + 32 + 42 +…+ n2 = n(n+1)(n+2)/6Let the given statement be P(n), i.e.,P(n)= 12 + 22 + 32 + 42 +…+ n2 = n(n+1)(2n+1)/6For n = 1,P(1) : 1 = 1(1+1)(2x1+1)/6 = 1x2x3/6 = 6/6 = 1, which is true.Let us assume that P(k) is true for n = k,P(k) = 12 + 22 + 32 + 42 +…+ k2 = k(k+1)(k+2)/6We shall now prove that P(k + 1) is also true. Now, we haveP(k) = 12 + 22 + 32 + 42 +…+ k2 + (k+1)2
= {k(k+1)(k+2)/6} + (k+1)2
= {k(k+1)(k+2) + 6(k+1)2}/6={(k+1)(2k2+7k+6)}/6={(k+1)(k+1+1)[2(k+1)+1]}/6= RHS
SOLCING THE INEQUALITIES
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XI-C32
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