Prove the following by using the principle of mathematical induction for all n ∈ N:
1/(2.5) . 1/(5.8) . 1/(8.11) . ..... . 1/[(3n - 1)(3n + 2)] = n/(6n + 4)

Solution:

Let P (n) be the given statement.

i.e., P (n) : 1/(2.5) . 1/(5.8) . 1/(8.11) . ..... . 1/[(3n - 1)(3n + 2)] = n/(6n + 4)

For n = 1,

P (1) : 1/(2.5) = 1/(6.1 + 4)

1/10 = 1/10, which is true.

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

i.e., 1/(2.5) . 1/(5.8) . 1/(8.11) . ..... . 1/[(3k - 1)(3k + 2)] = k/(6k + 4) ....(1)

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

Now, we have

1/(2.5) . 1/(5.8) . 1/(8.11) . ..... . 1/[3(k + 1) - 1] [3(k + 1) + 2]

= 1/(2.5) . 1/(5.8) . 1/(8.11) . ..... . 1/[(3k - 1)(3k + 2)] + 1/[3(k + 2)(3k + 5)]

= k/(6k + 4) + 1/[3(k + 2)(3k + 5)] ....[from (1)]

= 1 / (3k + 2) [k/2 + 1/(3k + 5)]

= 1 / (3k + 2)  [k(3k + 5) + 2 / (2(3k + 5)) ]

= 1 / (3k + 2) [(3k² + 5k + 2) / (2(3k + 5)) ]

= 1 / (3k + 2) [3k² + 3k + 2k + 2) / (2(3k + 5)) ]

= 1 / (3k + 2) [3k(k + 1) + 2 (k + 1)] / (2(3k + 5)) ]

= 1 / (3k + 2) [(k + 1)(3k + 2)] / 2(3k + 5) ]

= (k + 1)/(6k + 10)

= (k + 1)/[(6k + 6) + 4]

= (k + 1)/[6(k + 1) + 4]

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 10

Prove the following by using the principle of mathematical induction for all n ∈ N: 1/(2.5) . 1/(5.8) . 1/(8.11) . ..... . 1/[(3n - 1)(3n + 2)] = n/(6n + 4)

Summary:

We have proved that 1/(2.5) . 1/(5.8) . 1/(8.11) . ..... . 1/[(3n - 1)(3n + 2)] = n/(6n + 4) by using the principle of mathematical induction for all n ∈ N