Principle of Mahtematical Induction

1 )

For all n ∈ N – {1}, 7²ⁿ – 48n – 1 is divisible by ________

1234

2304

2348

2500

View Solution

Answer :

Answer: (d) 2304

P(n): 7²ⁿ– 48n – 1

P(1): 0 = 0 × 2304

P(2): 2304 = 1 × 2304

∴ P(1) & P(2) are true (1)

Let P(K) = 7^2k – 48K – 1 = m × 2304, m ∈ N and

P(K + 1) = 7^2(k+1) – 48(K + 1) – 1= m' × 2304, m' ∈ N be true (2)

P(K + 2) = 7^2k+4 – 48(K + 2) – 1 = 49 × 7^2K+2 – 48(K + 1) – 49

= 49 × 7^2k+2 – 48(K + 1) – 49

= 49(7^2k+2 – 1) – 48(K + 1)

= 49(2304m' + 48K + 48) – 48K – 48 ------ from (2)

= (49 × 2304m') + (49 × 48K) + (49 × 48) – (48K + 48)

= (49 × 2304m') + 48K(49 – 1) + 48(49 – 1)

= 49 × 2304m' + 48K × 48 + 48 + 48

= 2304 (49m' + K + 1)

= 2304 × m" where m" = 49m' + K + 1 is positive integer

= 2304

2 )

∀ n ∈ N, P(n): 2.7ⁿ + 3.5ⁿ – 5 is divisible by _______

View Solution

Answer :

P(n): 2.7ⁿ + 3.5ⁿ – 5

P(1): (2 × 7) + (3 × 5) – 5 = 24

P(2): 98 + 75 – 5 = 168 = 7 × 24

P(3): 686 + 375 – 5 = 44 × 24

∴ P(n) is divisible by 24

3 )

For all n ≥ 2, n² (n⁴ – 1) is divisible by _______

View Solution

Answer :

P(n): n² (n⁴ – 1) ------ for n ≥ 2

∴ P(2): 4(16 – 1) = 60,

P(3): 9 × (81 – 1) = 720 = 60 × 12

∴ From options, it is divisible by 60

4 )

For n ∈ N 10^n–2 > 81n, if _______

n > 5

n ≥ 5

n < 5

n > 6

View Solution

Answer :

For n ∈ N, 10^n–2 >81 n i.e. P(n): 10^n–2 > 81 n

P(0): 10^–2> 0 ….. not valid

P(1): 10^–1 > 81 ….. not valid

P(2): 1 > 162 ….. not valid

P(4): 100 > 81 × 4 ….. not valid

P(5): 1000 > 405 ….. This is true

P(6): 10000 > 486 ….. This is true

∴ n ≥ 5

5 )

For each n ∈ N, the correct statement is _______

2^n < n

n^2 > 2^n

n^4 < 10^n

2^(3n) > 7n +1

View Solution

Answer :

Consider options, for n = 1, 2¹ < 1 not true

1² > 2¹ not true

1⁴ < 10¹true

2³ > 7 + 1 not true

Hence option c is valid i.e. n⁴ < 10ⁿ

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