## Kerala Plus One Maths Chapter Wise Previous Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 1.

If 3^{2n+2} – 8n – 9 is divisible by ‘k’ for all n∈N is true, then which one of the following is a value of ‘k’ ?

i. 8

ii. 6

iii. 3

iv. 12

b. Prove by using the principle of Mathematical Induction

is true for all n∈N. [March-2018]

Answer:

a. 8

Question 2.

Consider the statement “10^{2n–}^{1}+1 is divisible by 11”. Verify that P (1) is true and then prove the statement by using Mathematical induction. [March-2017]

Answer:

Question 3.

Consider the following statement:

P(n): a + ar + ar^{2} +………… + ar ^{n-1 }=

a. Prove that P(l) is true.

b. Hence by using the principle of mathematical induction, prove that P(n) is true for all natural numbers n. [March-2016]

Answer:

Question 4.

A statement p (n) for a natural number n is given by

a. Verify that p (1) is true.

b. By assuming that p (k) is true for a natural number k, show that p (k+1) is true. [March-2015]

Answer:

Question 5.

Consider the statement “3^{2}^{n+}^{2 }-8n-9 by 8″.

a. Verify the statement for n = 1.

b. Prove the statement using the principle of mathematical induction for all natural numbers.

[March-2014]

Answer:

Question 6.

Consider the statement P(n) = 1 + 3 + 3^{2} + …+3^{n-1 }**= **

a. Check whether P(l) is true.

b. If P (k) is true, prove that P (k+1) is also true.

c. Is P (n) true for all natural numbers n? Justify your answer. [March-2013]

Answer:

a. P (1) = LHS = 1

P(1) =RHS=

P(1) [LHS]=p(1)[RHS]

∴ P(1) is True

c. Assume P (k) is true, we prove that P (k +1) is true, Hence by PMI P (n) is true for all natural numbers.

Question 7.

Consider the statement: “P(n): 9^{n} – 1 is a multiple of 8”

i. Is P(1) true?

ii. Assuming P (k) is true, prove that P(k +1) is true. [February-2013]

Answer:

Question 8.

Consider the statement: “n(n + 1)(2n +1) is divisible by 6”

a. Verify the statement for n = 2.

b. By assuming that P (k) is true for a natural number k, verify that P (k+1) is true. [March-2012]

Answer:

a. When n = 2,

2(2+ 1)(2×2+ 1) = 2×3×5 = 30 is divisible by 6.

b. P (k) = k (k + 1) (2k + 1) is divisible by 6

P(k+ 1)=(k+ 1)(k+2)(2k+ 1)+(k+ 1)(k+2)2

= (k+ 1)(2k+1)k+(k+ 1)(2k+ 1)x2+(k+1)(k+2)

= k(k+ 1)(2k+ 1) + 2(k+ 1)[2k+1 +k + 2]

= k(k+ 1)(2k+ 1) + 6(k+1)^{2} is divisible by 6

∴ P(k+ 1) is true.

Question 9.

Consider the statement “P(n):9^{n}-1, is a multiple of 8”, where n is a natural number.

i. Is P(1) true?

ii. Assuming P (k) is true, show that P (k + 1) is true. [March-2011]

Answer:

i. P(1) = 8 is true.P(k) ⇒ 9^{k} -1 is a multiple of 8

ii. P(k+ 1) ⇒ 9^{k+1}= 9^{k+1} -9 + 9-1

= 9 (9^{k-1}) + 8 is a multiple of 8.

Question 10.

Consider the statement “7^{n} – 3^{n} is divisible by 4”

i. Verify the result for n = 2.

ii. Prove the statement using mathematical

induction. [March-2010, August-2009]

Answer:

Question 11.

a. Which among the following is the least number that will divided 7^{2n} – 4^{2n} for every positive integer n?

[4, 7, 11, 33]

b. Prove by mathematical induction (cosnθ+ sinnθ)^{n}=(cosnθ+sinnθ) where [September-2010]

Answer:

a. 11

Question 12.

a. For every positive integer n, 7^{n} – 3^{n} should be divisible by

(2, 3, 4, 8)

b. Motion that 2 + 22 + 23 + 24 +… + 2^{n}= 2(2^{n} -1) [March-2009]

Answer:

Question 13.

Let P (n): 1 + 3 + 5 +… + (2n -1) = n2

i. Verify P (1) is true.

ii. Write P (k).

iii.Verify P (k +1) is true when P (k) is true. [June, February-2008]

Answer:

i.P(1): 1 =1^{2} P(l) is true,

since LHS = RHS=1

P(k): 1 + 3 + 5 + ….+ (2k- 1) = k^{2 }P (k + 1) : 1+3 + 5 +…………. ( + (2k – 1) + [2 (k + 1)-1] — (k+ 1)^{2 }Now 1 +3 + 5 +….. + (2k-1)+ [2 (k+1) -1]

= k^{2} + 2k + 1 =(k + 1)^{2 }P(k+ 1) is true

Question 14.

Let P(n) be the statement “3^{2}“ -1 is divisible by 8”

i. Is P(1) true? Justify your answer.

ii. If P (k) is true, show that P (k+1) is also true. [June-2007]

Answer:

i. Yes. P (1) is true.

P (1) : 3^{2} -1 is divisible by 8

3^{2} -1 =9-1 = 8, which is divisible by 8

ii. Let P (k): 3^{2k} -1 is divisible by 8 be true.

To show that P (k + 1) is true.

P (k + 1) : 3^{2(k+!)} -1 is divisible by 8

3^{2(k+1)}-1 =3^{2k}3^{2 }-1=3^{2k}(8-1)-1

= 8.3^{2k}+(3^{2k} -1)

The first term is divisible by 8 and second is also divisible by 8 (from P (k))

⇒ 3^{2(k+1)} -1 is divisible by 8

⇒ P(k+ 1) is true

Question 15.

P (n) is the statement: “n^{2} – n + 7 is a prime number”

i. Verify that P(l), P(2), P(3) and P(4) are true.

ii. Can you say that P (n) will be true for all natural numbers? Give reason. [February-2007]

Answer:

i. P(1) =1^{2}-1+7 (prime)

P(2) = 2^{2}– 2 + 7 = 9 (not prime)

P(3) = 3^{2}– 3 + 7 = 13 (prime)

P(4) = 4^{2}– 4 + 7 = 19 (prime)

P (1), P (3), P (4) are true P (2) is not true.

ii. No. P (n) is not true y n∈N

Question 16.

i. The sum of the first n natural numbers is

ii. Prove that the above result by the principle of mathematical induction for every natural number n.

[June-2006]

Answer: