Prove the following by using the principle of mathematical induction for all n ∈ N:
1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + .... + 1/ [n (n + 1)(n + 2)] = [n (n + 3)] / [4(n + 1)(n + 2)]

Solution:

Let P (n) be the given statement.

i.e., P (n) : 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + .... + 1/ [n (n + 1)(n + 2)] = [n (n + 3)] / [4(n + 1)(n + 2)]

For n = 1,

P (1) : 1/(1.2.3) = 1 (1 + 3)/4(1 + 1)(1 + 2)

1/6 = 1.4/4.2.3

1/6 = 1/6, which is true.

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

i.e., 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + .... + 1/ [k (k + 1)(k + 2)] = [k (k + 3)] / [4(k + 1)(k + 2)] ....(1)

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

Now, we have

LHS = 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + .... + 1/[(k + 1) [(k + 1) + 1] [(k + 1) + 2] ]

=1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + .... + 1/ [k (k + 1)(k + 2)] + 1/[(k + 1)(k + 2)(k + 3)]

= [k (k + 3)] / [4 (k + 1)(k + 2)] + 1/[(k + 1)(k + 2)(k + 3)] ....[from (1)]

= 1/[(k + 1)(k + 2)] [k (k + 3)/4 + 1/(k + 3)]

= 1/[(k + 1)(k + 2)] [(k (k + 3)² + 4) / (4 (k + 3)) ]

= 1/[(k + 1)(k + 2)] [(k³ + 6k² + 9k + 4) / (4 (k + 3))] ... (2)

RHS = (k + 1) [(k + 1) + 3] / [4(k + 1 + 1) (k + 1 + 2) ]

= [(k + 1) (k + 4)] / [4(k + 2) (k + 3)]

multiplying and dividing by (k + 1),

= 1/[(k + 1)(k + 2)] [(k + 4)(k + 1)²/4 (k + 3)]

= 1/[(k + 1)(k + 2)] [(k + 4) (k² + 2k + 1)/4 (k + 3)]

= 1/[(k + 1)(k + 2)] [(k³ + 2k² + k + 4k² + 8k + 4)/4 (k + 3)]

= 1/[(k + 1)(k + 2)] [(k³ + 6k² + 9k + 4) / (4 (k + 3))] ... (3)

From (2) and (3), LHS = RHS.

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 11

Prove the following by using the principle of mathematical induction for all n ∈ N: 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + .... + 1/ [n (n + 1)(n + 2)] = [n (n + 3)] / [4(n + 1)(n + 2)]

Summary:

We have proved that 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + .... + 1/ [n (n + 1)(n + 2)] = [n (n + 3)] / [4(n + 1)(n + 2)] by using the principle of mathematical induction for all n ∈ N