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Explore popular questions from Principle of Mathematical Induction for JEE Main. This collection covers Principle of Mathematical Induction previous year JEE Main questions hand picked by experienced teachers.

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Principle of Mathematical Induction

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Q 1. This question contains two statements : Statement-1 (Assertion) and Statement- 2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.
Statement-1: For every natural number {tex} n \geq 2 {/tex}, {tex} \frac { 1 } { \sqrt { 1 } } + \frac { 1 } { \sqrt { 2 } } + \ldots + \frac { 1 } { \sqrt { n } } > \sqrt { n } {/tex}
Statement-2 : For every natural number {tex} n \geq 2 {/tex}, {tex} \sqrt { n ( n + 1 ) } < n + 1 {/tex}

A

Statement-1 is true, Statement- 2 is false

B

Statemen-1 is false, Statement- 2 is true

C

Statement-1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1

Statement-1 is true, Statement- 2 is true; Statement- 2 is not a correct explanation for Statement-1

Explanation





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Q 2. Let {tex} S ( k ) = 1 + 3 + 5 + \ldots + ( 2 k - 1 ) = 3 + k ^ { 2 } {/tex}. Then which of the following is true?

A

{tex} S ( k ) \Rightarrow S ( k - 1 ) {/tex}

{tex} S ( k ) \Rightarrow S ( k + 1 ) {/tex}

C

{tex} S ( 1 ) {/tex} is correct

D

principle of mathematical induction can be used to prove the formula.

Explanation


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Q 3. If {tex} a _ { n } = \sqrt { 7 + \sqrt { 7 + \sqrt { 7 + \ldots . . } } } {/tex} having {tex} n {/tex} radical signs then by methods of mathematical induction which is true

A

{tex} a _ { n } > 7 , \forall \ n \geq 1 {/tex}

{tex} a _ { n } > 3 , \forall\ n \geq 1 {/tex}

C

{tex} a _ { n } < 4 , \forall \ n \geq 1 {/tex}

D

{tex} a _ { n } < 3 , \forall \ n \geq 1 {/tex}

Explanation


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Q 4. The greatest positive integer. which divides {tex} ( \mathrm { n } + 16 ) ( \mathrm { n } + 17 ) ( \mathrm { n } + 18 ) ( \mathrm { n } + 19 ) , {/tex} for all {tex} \mathrm { n } \in \mathrm { N } , {/tex} is-

A

{tex}2{/tex}

B

{tex}4{/tex}

{tex}24{/tex}

D

{tex}120{/tex}

Explanation



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Q 5. Let {tex} P ( n ): n ^ { 2 } + n {/tex} is an odd integer. It is seen that truth of {tex} P ( n ) \Rightarrow {/tex} the truth of {tex} P ( n + 1 ) . {/tex} Therefore, {tex} P ( n ) {/tex} is true for all-

A

{tex} n > 1 {/tex}

B

{tex} n {/tex}

C

{tex} n > 2 {/tex}

None of these

Explanation



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Q 6. If {tex} \mathrm { n } \in \mathrm { N } , {/tex} then {tex} \mathrm { x } ^ { 2 \mathrm { n } - 1 } + \mathrm { y } ^ { 2 \mathrm { n } - 1 } {/tex} is divisible by-

{tex} x + y {/tex}

B

{tex} x - y {/tex}

C

{tex} x ^ { 2 } + y ^ { 2 } {/tex}

D

{tex} x ^ { 2 } + x y {/tex}

Explanation

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Q 7. {tex} 1.2 ^ { 2 } + 2.3 ^ { 2 } + 3.4 ^ { 2 } + \ldots . {/tex} upto {tex} \mathrm { n } {/tex} terms, is equal to-

A

{tex}\mathrm{ \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( n + 3 ) }{/tex}

B

{tex} \frac { 1 } { 12 } \mathrm { n } ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) ( \mathrm { n } + 5 ) {/tex}

{tex}\mathrm{\frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + 5 ) }{/tex}

D

None of these

Explanation

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Q 8. The sum of the cubes of three consecutive natural numbers is divisible by-

A

2

B

5

C

7

9

Explanation

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Q 9. If {tex} \mathrm { n } \in \mathrm { N } , {/tex} then {tex} 11 ^ { \mathrm { n } + { 2 } } + 12 ^ { 2 \mathrm { n } + 1 } {/tex} is divisible by

A

113

B

123

133

D

None of these

Explanation

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Q 10. If {tex} \mathrm { n } \in \mathrm { N } , {/tex} then {tex} 3 ^ { 4 \mathrm { n } + 2 } + 5 ^ { 2 \mathrm { n } + 1 } {/tex} is a multiple of-

{tex}14{/tex}

B

{tex}16{/tex}

C

{tex}18{/tex}

D

{tex}20{/tex}

Explanation

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Q 11. The difference between an +ve integer and its cube is divisible by-

A

4

6

C

9

D

None of these

Explanation

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Q 12. For every positive integer
{tex} n , \frac { n ^ { 7 } } { 7 } + \frac { n ^ { 5 } } { 5 } + \frac { 2 n ^ { 3 } } { 3 } - \frac { n } { 105 } {/tex} is -

an integer

B

a rational number

C

a negative real number

D

an odd integer

Explanation

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Q 13. The sum of n terms of the series
{tex} \frac { \frac { 1 } { 2 } \cdot \frac { 2 } { 2 } } { 1 ^ { 3 } } + \frac { \frac { 2 } { 2 } \cdot \frac { 3 } { 2 } } { 1 ^ { 3 } + 2 ^ { 3 } } + \frac { \frac { 3 } { 2 } \cdot \frac { 4 } { 2 } } { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } } + \ldots \ldots {/tex}

A

{tex} \frac { 1 } { n ( n + 1 ) } {/tex}

{tex} \frac { n } { n + 1 } {/tex}

C

{tex} \frac { n + 1 } { n } {/tex}

D

{tex} \frac { n + 1 } { n + 2 } {/tex}

Explanation


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Q 14. For all {tex} \mathrm { n } \in \mathrm { N } , 7 ^ { 2 \mathrm n } - 48 \mathrm { n } - 1 {/tex} is divisible by {tex} - {/tex}

A

25

B

26

C

1234

2304

Explanation

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Q 15. The {tex}{ n } ^ { \text {th } } {/tex} term of the series
{tex} 4 + 14 + 30 + 52 + 80 + 114 + \ldots \ldots {/tex} is-

A

{tex} 5 n - 1 {/tex}

B

{tex} 2 n ^ { 2 } + 2 n {/tex}

{tex} 3 n ^ { 2 } + n {/tex}

D

{tex} 2 n ^ { 2 } + 2 {/tex}

Explanation

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Q 16. If {tex} 10 ^ { n } + 3.4 ^ { n + 2 } + \lambda {/tex} is exactly divisible by 9 for all {tex} n \in N , {/tex} then the least positive integral value of {tex} \lambda {/tex} is-

5

B

3

C

7

D

1

Explanation

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Q 17. The sum of {tex} n {/tex} terms of the series {tex} 1 + ( 1 + a ) + ( 1 + a + a ) + \left( 1 + a + a ^ { 2 } + a ^ { 3 } \right) + \ldots {/tex} is-

{tex} \frac { \mathrm { n } } { 1 - \mathrm { a } } - \frac { \mathrm { a } \left( 1 - \mathrm { a } ^ { \mathrm { n } } \right) } { ( 1 - \mathrm { a } ) ^ { 2 } } {/tex}

B

{tex} \frac { \mathrm { n } } { 1 - \mathrm { a } } + \frac { \mathrm { a } \left( 1 - \mathrm { a } ^ { \mathrm { n } } \right) } { ( 1 - \mathrm { a } ) ^ { 2 } } {/tex}

C

{tex} \frac { \mathrm { n } } { 1 - \mathrm { a } } + \frac { \mathrm { a } \left( 1 + \mathrm { a } ^ { \mathrm { n } } \right) } { ( 1 - \mathrm { a } ) ^ { 2 } } {/tex}

D

{tex} - \frac { \mathrm { n } } { 1 - \mathrm { a } } + \frac { \mathrm { a } \left( 1 - \mathrm { a } ^ { \mathrm { n } } \right) } { ( 1 - \mathrm { a } ) ^ { 2 } } {/tex}

Explanation

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Q 18. For all {tex} { n } \in \mathrm { N } , \cos \theta \cos 2 \theta \cos 4 \theta \ldots \ldots \cos 2 ^ { n - 1 } \theta {/tex} equals to-

{tex} \frac { \sin 2 ^ { n } \theta } { 2 ^ { n } \sin \theta } {/tex}

B

{tex} \frac { \sin 2 ^ { n } \theta } { \sin \theta } {/tex}

C

{tex} \frac { \cos 2 ^ { n } \theta } { 2 ^ { n } \cos 2 \theta } {/tex}

D

{tex} \frac { \cos 2 ^ { n } \theta } { 2 ^ { n } \sin \theta } {/tex}

Explanation

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Q 19. The smallest positive integer for which the statement {tex} 3 ^ { n + 1 } < 4 ^ { n } {/tex} holds is-

A

1

B

2

C

3

4

Explanation

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Q 20. If {tex} P {/tex} is a prime number then {tex} n ^ { p } - n {/tex} is divisible by p when {tex} n {/tex} is a

natural number greater than 1

B

odd number

C

even number

D

None of these

Explanation

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Q 21. For all positive integral values of n, 32n − 2n + 1 is divisible by

2

B

4

C

8

D

12

Explanation

On putting n = 2 in 32n − 2n + 1, we get

32 × 2 − 2 × 2 + 1 = 81 − 4 + 1 = 78

Which is divisible by 2

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Q 22. 3 + 13 + 29 + 51 + 79 + … to n terms =

A

2n2 + 7n3

B

n2 + 5n3

n3 + 2n2

D

None of these

Explanation

Clearly, n3 + 2n2 gives the sum of the series for n = 1, 2, 3 etc.

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Q 23. $\text{If\ }A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}\ \text{and\ }I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix},$ then which one of the following holds for all n ≥ 1, by the principle of mathematical induction?

A

An = 2n − 1A + (n−1)I

B

An = nA + (n−1)I

C

An = 2n − 1A − (n−1)I

An = nA − (n−1)I

Explanation

$A^{2} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}$

$A^{3} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \\ \end{bmatrix}$

$A^{n} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}$ can be verified by induction. Now, taking option

$\left( b \right)\begin{bmatrix} 1 & 0 \\ n & 1 \\ \end{bmatrix} = \begin{bmatrix} n & 0 \\ n & n \\ \end{bmatrix} + \begin{bmatrix} n - 1 & 0 \\ 0 & n - 1 \\ \end{bmatrix}$

$\Longrightarrow \begin{bmatrix} 1 & 0 \\ n & 1 \\ \end{bmatrix} \neq \begin{bmatrix} 2n - 1 & 0 \\ 1 & 2n - 1 \\ \end{bmatrix}$

$\left( d \right)\text{nA} - \left( n - 1 \right)I = \begin{bmatrix} n & 0 \\ n & n \\ \end{bmatrix} - \begin{bmatrix} n - 1 & 0 \\ 1 & n - 1 \\ \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ n & 1 \\ \end{bmatrix}{= A}^{n}$

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Q 24. If (n) : 1 + 3 + 5 + ... + (2n−1) = n2 is

True for all n ∈ N

B

True for n > 1

C

True for no n

D

None of these

Explanation

Given, (n) : 1 + 3 + 5 + ... + (2n−1) = n2

P(1) : 1 = 1(true)

Let P(k) = 1 + 3 + 5 + ... + (2k−1) = k2

∴ P(k+1) = 1 + 3 + 5 + ... + (2k−1) + 2k + 1

= k2 + 2k + 1 = (k+1)2

So, it holds for all n.

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Q 25. For a positive integer n, Let $a\left( n \right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{\left( 2^{n} \right) - 1}.$ Then

a(100) ≤ 100

B

a(100) > 100

C

a(200) ≤ 100

D

None of these

Explanation

It can be proved with the help of mathematical induction that $\frac{n}{2} < a(n) \leq n$.

$\therefore\frac{200}{2} < a\left( 200 \right) \Rightarrow a\left( 200 \right) > 100$

and a(100) ≤ 100