Important Questions Class 11 Maths Chapter 4

Important Questions Class 11 Mathematics Chapter 4

Important Questions for CBSE Class 11 Mathematics Chapter 4 – Principle of Mathematical Induction

Important Questions Class 11 Mathematics Chapter 4 prepared by Extramarks, will help students gain a better understanding of concepts and terminologies related to the chapter, Principle of Mathematical  Induction. Students can get answers to their doubts as these notes are comprehensive. These notes are compiled by subject matter experts to give a brief overview of all the important questions in this chapter. Chapter 4 Class 11 Mathematics Important Questions are made in accordance with the CBSE Syllabus. 

These notes are beneficial for students while preparing for the exams, as they cover all the main topics of this chapter. Students can go through these notes before their exams so that they can brush up on all the important formulas and concepts.

CBSE Class 11 Mathematics Chapter 4 Important Questions

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4 Marks Answers and Questions

Q1. Prove that 102n-1+1 is divisible by 11.

Ans. Using the method of P.M.I.,

       Given, P(n):102n-1+1

Checking if the statement is true or not for n=1

For n=1

P(1):10(21)-1+1=11.

Which is divisible by 11.

Therefore, P(1) is true.

Let, P(n):102n-1+1 is true for n=k.

Q2. Prove 2n+7<n+32.

Ans. Using the method of P.M.I.,

Given, P(n):2n+7<n+32.

Checking if the statement is true or not for n=1.

So, for n=1.

P(1): 9<42

9<16

which is true.

Thus, P(1) is true.

Let, P(n) : 2n+7<n+32  is true for n=k.

That is, P(k):2k+7<k+32                                     …………(1)

Now, we have to show that the given statement

 P(n):2(k+1)+7<k+1+32 is true for n=k+1

So, P(k+1):(2k+9)<k+42

Now, (2k+7+2)<k+42

k+32+2<k+42 from Equation (1)

k2+9+6k+2<k+42

k2+6k+11<k+42

k2+6k+11+2k-2k+5-5<k+42

k2+8k+16–2k+5<k+42

k+42–2k+5<k+42

Which is true.

Hence, P(k+1):(2k+9)<k+42 is true.

Thus, P(k+1) is true when Pk is true.

Therefore, by P.M.I. the statement (2n+7)<n+32  is true for all n N.

Q3. Using induction, prove that 10n+34n+2+5 is divisible by 9.

Ans. 10n+34n+2+5 is divisible by 9 by using the method of PMI.

Given, P(n): 10n+34n+2+5.

Checking if the statement is true or not for n=1.

So, For n=1.

P(1):101+341+2+5=207.

Which is divisible by 9.

Thus, P(1) is true.

Let, P(n):10n+34n+2+5 is true for n=k.

That is, P(k): 10k+34k+2+5=9 , where N                                           …..(1)

Now, we have to show that the given statement P(n):10n+34n+2+5 is true for n=k+1 

P(k+1):10k+1+34k+1+2+5

10k+1+3 4k+1+2+5

10k10+3 4k43+5

(9 -34k+2-5)10+34k43+5 {from equation(1)}

90 -304k+2-50+34k43+5

90 -304k+2-45+34k+24

90 – 304k+2-45+124k+2

90 – 184k+2-45

910 -24k+2-5

Hence, P(k+1):10k+1+34k+1+2+5 is divisible by 9.

Thus, P(k+1) is true when P(k) is true.

Therefore, by P.M.I. the given statement is true for every positive integer n.

Q4. Prove that n(n+1)(n+5) is a multiple of 3.

Ans: nn+1n+5 is a multiple of 3 by using the method of PMI.

Given, P(n):nn+1n+5

Checking if the statement is true or not for n=1.

So, For n=1

P(1):11+11+5=12

Which is multiple of 3.

Thus, P(1) is true.

Let, P(n):nn+1n+5 is true for n=k.

That is, P(k):kk+1k+5=3 , where N                                                      ……(1)

Now, we have to show that the given statement P(n):nn+1n+5 is true for n=k+1. 

P(k+1):(k+1)(k+2)(k+6)

(k+1)(k+2)(k+6)=[(k+1)(k+2)](k+6)

k(k+1)(k+2)+6(k+1)(k+2)

k(k+1)(k+5-3)+6(k+1)(k+2)

k(k+1)(k+5)+3k(k+1) +6(k+1)(k+2)

k(k+1)(k+5)+ (k+1) 6(k+2)-3k

k(k+1)(k+5)+ (k+1)3k+12

k(k+1)(k+5)+ 3k+1k+4

3 + 3k+1k+4 from equation (1)

3 +k+1k+4

Hence, P(k+1):k+1k+2k+6 is multiple of 3.

Thus, P(k+1) is true when P(k) is true.

Therefore, by P.M.I. the given statement n(n+1)(n+5) is multiple of 3.

Q5. Show that the sum of the first n odd natural number is n2.

Ans. Using the PMI method to prove 1+3+5+…..+(2n-1)=n2

Given,  P(n):1+3+5+…..+(2n-1)=n2

Checking if the statement is true or not for n=1.

So, For n=1

P(1):1=1

Which is true

Thus, P(1) is true.

Let P(n):1+3+5+……+(2n-1)=n2  is true for n=k.

That is, P(k):1+3+5+……+(2k-1)=k2    …………………….(1)

Now, we have to show that the given statement 

P(n):1+3+5+……+(2n-1)=n2 is true for n=k+1

So,  P(k+1):1+3+5+……+(2(k+1)-1)=(k+1)2

 P(k+1):1+3+5+……+(2k+1)=(k+1)2

Now, L.H.S =1+3+5+……+(2k-1)+(2k+1)

=k2+(2k+1) …………………..{from equation} 

(k+1)2=   R.H.S

Which is true.

Hence, P(k+1):1+3+5+……+(2(k+1)-1)=(k+1)2  is true.

Thus, P(k+1) is true when P(k) is true.

Therefore, by P.M.I. the statement 1+3+5+……+(2n-1)=n2  is true.

Q6. Prove x2n-1 is divisible by x-1.

Ans. x2n-1 is divisible by x-1 with the method of PMI.

Given, P(n):x2n-1.

Checking if the statement is true or not for n=1.

So, For n=1

P(1):x2-1=(x-1)(x+1)

Which is divisible by (x-1).

Thus, P(1) is true.

Let, P(n):x2n-1 is true for n=k.

That is, P(k):x2k-1=x-1 , where N                                                          ………(1)

Now, we have to show that the given statement P(n):x2n-1 is true for n=k+1.

P(k+1):x2k+2-1

=x2kx2-1

={[(x-1) +1]x2-1} from equation (1)

=  {(x-1) x2 +x2-1}

=(x-1) x2+(x-1)(x+1)

=(x-1) x2+(x+1)

Hence, P(k+1):x2k+2-1 is divisible by (x-1).

Thus, P(k+1) is true when P(k) is true.

Therefore, by P.M.I. the given statement is true for every positive integer n . 

Q7. The sum of the cubes of three consecutive natural numbers is divisible by 9.

Ans. [n3+n+13+n+23].

Checking if the statement is true or not for n=1.

So, For n=1.

P(1):13+23+33=243

divisible by 9.

Thus, P(1) is true.

Let, P(n): [n3+n+13+n+23] is true for n=k.

P(k): [k3+k+13+k+23]=9 , where N                                          ……………1

Now, we have to show that the given statement P(n):[n3+n+13+n+23] is true for n=k+1.

P(k+1):[k+13+k+23+k+33]

=[k+13+k+23+k3+9k2+27k+27]

=9 +9k2+27k+27 from equation (1)

=9[ +k2+3k+3]

divisible by 9.

Hence, P(k+1):[k+13+k+23+k+33] is divisible by 9.

Thus, P(k+1) is true when Pk is true.

Therefore, by P.M.I. the given statement is true for every positive integer n.