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Q1. If P(n) be the statement `12n + 3’ is a multiple of 5, then show that P(3) is false, whereas P(6) is true.

Solution

P(n): `12n + 3’ is a multiple of 5. P(3) is the statement  `12 x 3 + 3’ is a multiple of 5, which is false. And P(6) is the statement  `12 x 6 + 3’ is a multiple of 5, which is true.

Q2. Prove by using the principle of mathematical induction thatx²ⁿ - y²ⁿ is divisible by (x + y).

Solution

Q3.

Solution

Q4. Let P(n) be the statement n³ + (n + 1)³ + (n + 2)³ is divisible by 9 for all n N. If P(k) is true, then show that P(k + 1) is true.

Solution

Q5. If  P(n) be the statement ‘n² – n + 41 is a prime number’, then show that P(1), P(2) are true but P(41) is not true.

Solution

P(n): n² – n + 41 is a prime number. P(1): 1² – 1 + 41 is a prime number i.e. P(1) = 41 is prime, which is true. P(2): 2² – 2 + 41 is a prime number  i.e. P(2) = 43 is prime, which is true. P(41): 41² – 41 + 41 is a prime number  i.e. P(41) = 41² is prime, which is not true.

Q6. Show by using principle of mathematical induction, prove that: 1 + 2 + 3 + -----+ n =

Solution

Let the given statement is P(n)   P(n) = 1 + 2 + 3 + --- + n =   For n = 1,   P(1) = is true, since 1 =   Assume that P(k) is true for some positive integer k, i.e. 1 + 2 + 3 + ---+ k = We shall prove that P(k +1) is true, whenever P(k) is true. We have, 1 + 2 + 3 + -----+ k + (k + 1) =   + (k + 1)                                                                Thus P(k +1) is true, whenever P(k) is true. Hence from principle of mathematical induction, the statement P(n) is true for all natural numbers N.

Q7. Prove that 10^{2n – 1} + 1 is divisible by 11 for all n N.

Solution

Let P(n): 10^{2n – 1} + 1 is divisible by 11 P(1): 10 + 1 = 11which is divisible by 11. Thus P(n) is true for n = 1. Let P(k) be true for some natural number k. i.e. 10^{2k – 1} + 1 is divisible by 11. Let 10^{2k – 1} + 1 = 11d Now we prove that P(k + 1) is true whenever P(k) is true. Now, P(k+1): 10^2k+1 + 1 = (11d – 1)100 + 1 = 11 (100d – 9),which is divisible by 11. P(k + 1) is true. Thus P(k +1) is true whenever P(k) is true. By principle mathematical induction 10^{2n – 1} + 1 is divisible by 11 for all n  N. 

Q8. Let P(n) be the statement . What are P(3) and P(n + 1)?

Solution

P(3): 3³ > 3 i.e. P(3): 27 > 3 P(n + 1): 3^{n + 1} > n + 1

Q9.

Solution

Q10. Let P(n) be the statement , If P (k) is true , then show that P(k + 1) is true.

Solution

We are given that P (k) is true. We shall prove that P (k +1) is true, whenever P (k) is true. P (k): k < 2^k Thus P(k + 1) is true, whenever P(k) is true.