# 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

Students can preview the Class 11 Mathematics Chapter 4 Important Questions given below. They can also click the link to view additional 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+162k+5<k+42

k+422k+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.

Q.1
Let P(n) be the statement ?n3 + n is divisible by 3?, then the incorrect statement is
P(3).
P(6).
P(9).
P(11).
Marks:1

Ans
P(11) is the statement 113 + 11 is divisible by 3 i.e.,

1342 is divisible by 3, that is not true

Therefore, the statement P(11) is false.
Q.2
10n + 3.22n + 4 + 5 (for n = 4) is divisible by
7.
9.
11.
13.
Marks:1

Ans
P(n) = 10n + 3.22n + 4 + 5

Now we check for the value of n = 4 and we get,

P(4) = 104 + 3.212 + 5

= 10000 + 12288 + 5

= 22293

= 9 × 2477

Therefore, we can observe that ?10n + 3.22n + 4 + 5 (for n = 4) is divisible by 9.
Q.3
Using the principle of mathematical induction,prove that13.7+17.11+111.15+ ? +1(4n-1)(4n+3)=n3(4n+3) for?all?n?N
Marks:4

Ans
Let the given statement bePn:13.7+17.11+111.15+14n-14n+3=n34n+3 .1Test for n=1LHS= 13.7=121RHS=n34n+3=134+3=121Hence,

P1 is true. Let Pn is true for n=k then 1 becomes13.7+17.11+111.15++14k-14k+3=k34k+3 2Now,

we shall test for n=k+11 becomes13.7+1711+111.15+14k+34k+7=k+134k+73 LHS=13.7+17.11+111.15+14k-14k+3+14k+34k+7=k34k+3+14k+34k+7

Using 2=14k+3k3+14k+7=14k+34k2+7k+334k+7 =14k+34k+3k+134k+7 =k+134k+7=RHSPk+1 is true. Hence, Pn is true for all nN.
Q.4
Prove that : 2n > n for all positive integers n.
Marks:3

Ans
When n=1,21>1. Hence P1 is TrueAssuming Pk is True Pk:2k>kWe shall prove Pk+1 is Truei.e.,

Pk+1=2k+1>k+1 Pk+1=2k+1=2k 21=2k 22k.2>2k=k+k>k+1 2k+1>2k>k+1Pk+1 is TrueHence by principle induction Pn is true, for all nN.
Q.5
Prove that: 2.7n + 3.5n ? 5 is divisible by 24.
Marks:6

Ans
Let P n=2.7n+3.5n-5P1 =2.71+3.51-5 =14+15-5 =24P1 is true Assuming Pk is truePk =2.7k+3.5k-5 is divisible by 24We shall prove Pk+1 is truei.e.

Pk+1=2.7k+1+3.5k+1-5 is divisible by 242.7k+1+3.5k+1-5=24g let 1Pk+1=2.7k+1+3.5k+1-5 =2.7k.7+3.5k.5-5

=72.7k+3.5k-5-3.5k+5+3.5k.5-5 =724g-3.5k-5-3.5k-5+15.5k-5 [From equation 1] =724g-21.5k+35+15.5k-5 =724g-6.5k+30 =724g-65K-5 =724g-6(4P) 5k-5

is a multiple of 45k-5=4P =724g-24 PWhich is divisible by 24 Pk+1 is true

Hence by principle of mathematical induction is true. For all n N

## Please register to view this section

### 1. What is ‘Mathematical Induction’? Why do we use it?

Mathematical Induction is a mathematical technique that is used to prove a statement, a formula, or a theorem is true for every natural number. Mathematical induction is usually used in case of all natural numbers to state whether a given statement holds true.

### 2. Name the types of mathematical induction.

The types of mathematical induction are as follows:

• First principle of mathematical induction
• Second principle of mathematical induction
• Second principle of mathematical induction(variation)

### 3. How is Important Questions Class 11 Mathematics Chapter 4 beneficial for Students?

Important Questions Class 11 Mathematics Chapter 4 made by subject matter of  Extramarks is very useful. It has many benefits such as:

• This questionnaire consists of important questions which are categorised on the basis of marks distribution.
• These questions are comprehensive and concise.
• These questions are constructed by keeping in mind the CBSE guidelines.
• Chapter 4 Class 11 Maths Important Questions cover all important types of questions that can come for the exams.
• These questions are best for revision before exams.

### 4. Prove that 1+xn1+nx, for all natural numbers n, where x＞-1.

Let P(n) be the given statement,

i.e., P(n): 1+xn1+nx, for x＞-1.

We note that P(n) is true when   n=1, since 1+x1+x for  x＞-1.

Assume that

P(k): 1+xk1+kx, x＞-1 is true.                                                                        … (1)

We want to prove that P(k+1) is true for x＞-1 whenever P(k) is true.                     … (2)

Consider the identity

1+xk= 1+xk1+x

Given that x＞-1, so 1+x > 0.

Therefore, by using 1+xk1+kx

we have 1+xk+11+kx1+x

i.e.1+xk+11+x+kx+kx2.                                                                                   … (3)

Here k is a natural number and x20 so that kx2 ≥ 0.

Therefore 1+x+kx+kx21+x+kx

and so we get 1+xk+11+x+kx

i.e. 1+xk+11+1+kx

Thus, the statement in (2) is established.

Hence, as per the principle of mathematical induction,  P(n) is true for all natural numbers.