Prove the following by using the principle of mathematical induction for all n ∈ N :
(1 + 1/1)(1 + 1/2)(1 + 1/3) .... (1 + 1/n) = (n + 1)

Solution:

Let P (n) be the given statement.

i.e., P (n) : (1 + 1/1)(1 + 1/2)(1 + 1/3) .... (1 + 1/n) = (n + 1)

For n = 1,

P (1) : (1 + 1/1) = 1 + 1

2 = 2, which is true.

Assume that P (k) is true for some positive integer k.

i.e., P (k) : (1 + 1/1)(1 + 1/2)(1 + 1/3) .... (1 + 1/k) = (k + 1) ..... (1)

We will now prove that P (k + 1) is also true.

Now, we have

(1 + 1/1)(1 + 1/2)(1 + 1/3) .... (1 + 1/(k + 1))

= (1 + 1/1)(1 + 1/2)(1 + 1/3) .... (1 + 1/k) (1 + 1 / (k + 1))

= (k + 1) ( 1 + 1 / (k+1))   (From (1))

= (k+1) [ (k+1+1) / (k+1)]

= k + 2

Thus P (k + 1) is true, whenever P (k) is true.

Hence, from the principle of mathematical induction, the statement P (n) is true for all natural numbers i.e., n ∈ N .

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NCERT Solutions Class 11 Maths Chapter 4 Exercise 4.1 Question 14

Prove the following by using the principle of mathematical induction for all n ∈ N : (1 + 1/1)(1 + 1/2)(1 + 1/3) .... (1 + 1/n) = (n + 1)

Summary:

We have proved that (1 + 1/1)(1 + 1/2)(1 + 1/3) .... (1 + 1/n) = (n + 1) by using the principle of mathematical induction for all n ∈ N.