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Mathematical Induction

Mathematical Induction

 

Certain results in algebra are devised with n number of terms in them where n (a positive integer) is a natural number. The technique which proves useful for proving such kind of results is called the principle of mathematical induction. The notation which is generally used for the proposition P which depends on a positive integer n is P (n) read as P of n.

  • First Principle of Mathematical Induction

The proposition P (n) can be proved by the principle of mathematical induction for all n ∈ N as discussed in the following three steps:

1. The first step is also termed as the basic step of induction in which we prove that the position P(n) is true for the smallest positive integer or the first natural number i.e. for n = 1. This is also called as the verification step.

2. The next step is the induction step in which the proposition is assumed to be true for some n = k ≥ 1, i.e. we assume P(k) to be true.

3. The last step is the generalization step. If P(k) is true, then the proposition is proved true for n = k+1 i.e. for the next positive integer or the next natural number.

So, in this step the motive is to prove that if P(k) is true then P(k+1) must also be true, i.e. P(k) ⇒P(k+1) is true.

At the end, the result is generalized by saying that since the proposition is true for n = k+1, then it must be true for n = k as well and hence the proposition will also be true for all n belonging to the set of natural numbers.

  • Second Principle of Mathematical Induction

1. The first step is the verification step in which the proposition P(n) is proved to be true for n = m, where m is some fixed integer.

2. In the next step which is the induction step, we assume that the proposition P(n) is true for n = m, m+1, m+2, … r.

3. The last step is again the generalization step. In this step, prove that the proposition P(n) is true for n = r +1. Thus, once it is proved true, the result is generalized by stating that since the proposition is true for n = r+1, then it must also be true for n = m, m+1, m+2, … r as assumed in the previous step. Hence, the proposition is true for all n ≥ m belonging to the set of natural numbers.

Example 1:

By mathematical induction,

 \frac{1}{1.2.3} + \frac{1}{2.3.4} + .... + \frac{1}{n(n+1)(n+2)}

is equal to

 (1) \frac{n(n+1)}{4(n+2)(n+3)}                                                                              (2) \frac{n(n+3)}{4(n+1)(n+2)}

(3) \frac{n(n+2)}{4(n+1)(n+3)}                                                                               (4) \frac{n(n+2)}{4(n+2)(n+3)}

Solution:

Let P(n) = \frac{1}{1.2.3} + \frac{1}{2.3.4} + .... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}

(i) For n = 1

L.H.S. = \frac{1}{1.2.3} = \frac{1}{6} and R.H.S. = \frac{1.(1+3)}{4(1+1)(1+2 )} = \frac{1}{6}

therefore, P(1) is true.

(ii) Let P(k) be true, then

P(k) = \frac{1}{1.2.3} + \frac{1}{2.3.4} + .... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}  …..... (A)

(iii) For n = k+1,

P(k+1) = \frac{1}{1.2.3} + \frac{1}{2.3.4} + .... + \frac{1}{k(k+1)(k+2)} + \frac{1}{k+1(k+2)(k+3)}

                      = \frac{k+1)(k+4)}{4(k+2)(k+3)}

L.H.S. = \frac{1}{1.2.3} + \frac{1}{2.3.4} + .... + \frac{1}{k(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)}

               = \frac{k(k+3))}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)}    …...... from (A)

               = \frac{k(k+3)^{2} + 4}{4(k+1)(k+2)(k+3)}

               = \frac{k^{3} + 6k^{2} + 9k + 4}{4(k+1)(k+2)(k+3)}

               = \frac{(k+1)^{2} (k + 4)}{4(k+1)(k+2)(k+3)}

               = \frac{(k+1) (k + 4)}{4(k+2)(k+3)} 

               = R.H.S.

Hence, P(k+1) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

The correct option is (b).

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