11 Mathematics Ncert Ch04 Principle of Mathematical Induction 4.1 Solutions

Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in

Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks

Class XI Mathematics NCERT Solutions

PRINCIPLE OF MATHEMATICAL INDUCTION

Exercise 4.1

Answers

1. Let ( ) ( )2 1 3 1P 1 3 3 .......... 32

n

nn

= + + + + =

For 1n = ( ) ( )13 1

P 1 12

= = 1 = 1

( )P 1 is true. Now, let ( )P n be true for n k=

( ) ( )2 1 3 1P 1 3 3 .......... 32

k

kk

= + + + + = .(i)

For 1n k= + ( ) ( )2 1 3 1P 1 1 3 3 .......... 3 3 32

k

k k kk

+ = + + + + + = + [Using eq. (i)]

( )13 1 2.3 3.3 1 3 1

P 12 2 2

k k k k

k+ + + = = =

( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

2. Let ( ) ( )2

3 3 3 3 1P 1 2 3 ..........2

n nn n

+ = + + + + =

For 1n = ( ) ( )2

1 1 1P 1 1

2

+ = =

1 = 1


( ) ( )2

3 3 3 3 1P 1 2 3 ..........2

k kk k

+ = + + + + =

.(i)

For 1n k= + ( ) ( ) ( ) ( )2

3 33 3 3 3 1P 1 1 2 3 .......... 1 12

k kk k k k

+ + = + + + + + + = + +

( ) ( )2

2P 1 1 1

4

kk k k

+ = + + +

= ( )

22 4 4

14

k kk

+ ++

= ( ) ( )2 21 2

4

k k+ +

( ) ( )( )2

1 2P 1

2

k kk

+ + + =




3. Let ( ) ( ) ( ) ( ) ( )1 1 1 2

P 1 ..........1 2 1 2 3 1 2 3 ...... 1

nn

n n= + + + + =

+ + + + + + + +

For 1n = ( ) 2 1P 1 11 1

= =+

1 = 1


( ) ( ) ( ) ( ) ( )1 1 1 2

P 1 ..........1 2 1 2 3 1 2 3 ...... 1

kk

k k= + + + + =

+ + + + + + + + .(i)

For 1n k= + ( ) ( ) ( ) ( )1 1 1

P 1 1 ..........1 2 1 2 3 1 2 3 ......

kk

+ = + + + ++ + + + + + +

( ) ( ) ( )

1 2 1

1 2 3 ...... 1 1 1 2 3 ...... 1

k

k k k+ = +

+ + + + + + + + + + + [Using (i)]

( ) ( )( )2 1

P 11 21

2

kk

k kk+ = +

+ ++ = ( )( )

2 2

1 1 2

k

k k k+

+ + + =

( ) ( )21 2 121 2 2

k k

k k k

+ +=

+ + +


4. Let ( ) ( ) ( ) ( ) ( )( )1 2 3P 1.2.3 2.3.4 .......... 1 24

n n n nn n n n

+ + += + + + + + =

For 1n = ( ) 1 2 3 4P 1 1 2 34

= = 6 = 6


( ) ( )( ) ( ) ( )( )1 2 3P 1.2.3 2.3.4 .......... 1 24

k k k kk k k k

+ + += + + + + + = (i)

For 1n k= + ( ) ( )( ) ( ) ( ) ( )P 1 1.2.3 2.3.4 .......... 1 2 1 2 3k k k k k k k+ = + + + + + + + + +

( ) ( )( ) ( ) ( )( )1 2 3 1 2 3

4

k k k kk k k

+ + += + + + + [Using eq. (i)]

( ) ( )( )( )P 1 1 2 3 14

kk k k k

+ = + + + +

= ( )( ) ( ) 41 2 34

kk k k

+ + + +



( ) ( )( ) ( )( )1 2 3 4P 14

k k k kk

+ + + ++ =


5. Let ( ) ( )1

2 3 2 1 3 3P 1.3 2.3 3.3 .......... .34

nn nn n

+ += + + + + =

For 1n = ( ) ( )1 12 1 1 3 3

P 1 1 34

+ += = 3 = 3


( ) ( )1

2 3 2 1 3 3P 1.3 2.3 3.3 .......... .34

kk kk k

+ += + + + + =

For 1n k= +

( ) ( ) ( ) ( )1

2 3 1 12 1 3 3P 1 1.3 2.3 3.3 .......... .3 1 3 1 34

kk k kkk k k k

++ + ++ = + + + + + + = + +

( ) ( ) ( )1

12 1 3 3P 1 1 34 4

kkkk k

+++ = + + + = 1 2 1 33 1

4 4k k k+

+ + +

( ) 1 12 1 4 4 3 6 3 3P 1 3 34 4 4 4

k kk k kk + + + + + + = + = +

( ) ( )13 .3 2 1 3

P 14 4

k kk

+ ++ = + =

( ) 22 1 3 34

kk ++ +


6. Let ( ) ( ) ( )( )1 2P 1.2 2.3 3.4 .......... 13

n n nn n n

+ + = + + + + + =

For 1n = ( ) ( ) ( ) ( )1 1 1 1 2P 1 1 1 13

+ += + = 2 = 2


( ) ( ) ( ) ( )1 2P 1.2. 2.3. .......... 13

k k kk k k

+ += + + + + = (i)

For 1n k= +



( ) ( ) ( )( ) ( ) ( ) ( )( )1 2P 1 1.2 2.3 .......... 1 1 2 1 23

k k kk k k k k k k

+ ++ = + + + + + + + == + + +

( ) ( )( )P 1 1 2 13

kk k k

+ = + + +

= ( )( ) 31 23

kk k

+ + +

= ( )( ) ( )1 2 3

3

k k k+ + +


7. Let ( ) ( ) ( ) ( )24 6 1

P 1.3 3.5 5.7 .......... 2 1 2 13

n n nn n n

+ = + + + + + =

For 1n = ( ) ( )( )( )21 4 1 6 1 1

P 1 2 1 1 2 1 13

+ = + = 3 = 3


( ) ( )( ) ( )24 6 1

P 1.3 3.5 5.7 .......... 2 1 2 13

k k kk k k

+ = + + + + + =

For 1n k= + ( ) ( )( ) ( ) ( )P 1 1.3 3.5 5.7 .......... 2 1 2 1 2 1 1 2 1 1k k k k k+ = + + + + + + + + +

= ( ) ( ) ( )

24 6 12 1 1 2 1 1

3

k k kk k

+ + + + +

( ) ( ) ( )3 24 6

P 1 2 1 2 33

k k kk k k

+ + = + + + = ( )3 2 24 6 3 4 8 3

3

k k k k k+ + + +

( )3 2 24 6 12 24 9

P 13

k k k k kk

+ + + ++ = = 3 24 18 23 9

3

k k k+ + + =

( )( )21 4 14 93

k k k+ + +


8. Let ( ) ( )2 3 1P 1.2 2.2 3.2 .......... .2 1 2 2n nn n n += + + + + = + For 1n = ( ) ( )1 1 1P 1 1 2 1 1 2 2+= = + 2 = 2 ( )P 1 is true. Now, let ( )P n be true for n k= ( ) ( )2 3 1P 1.2 2.2 3.2 .......... .2 1 2 2k kk k k += + + + + = + For 1n k= +



( ) ( ) ( ) ( )2 3 1 1 1P 1 1.2 2.2 3.2 .......... .2 1 .2 1 2 2 1 .2k k k kk k k k k+ + ++ = + + + + + + = + + + ( ) ( ) ( )1 1P 1 1 .2 2 1 .2k kk k k+ ++ = + + + = ( )12 1 1 2k k k+ + + + = 12 2 2k k+ + = 2.2 2kk + + ( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

9. Let ( ) 1 1 1 1 1P .......... 12 4 8 2 2n n

n = + + + + =

For 1n = ( ) 1 11 1

P 1 12 2

= = 1 12 2

=


( ) 1 1 1 1 1P .......... 12 4 8 2 2k k

k = + + + + =

For 1n k= +

( ) 1 11 1 1 1 1 1 1

P 1 .......... 12 4 8 2 2 2 2k k k k

k + ++ = + + + + + = +

( ) 11 1

P 1 12 2k k

k + + =

= 1

2 11

2k +

=

1

11

2k +


10. Let ( ) ( ) ( ) ( )1 1 1 1

P ..........2.5 5.8 8.11 3 1 3 2 6 4

nn

n n n= + + + + =

+ +

For 1n = ( ) ( )( ) ( )1 1

P 13 1 1 3 1 2 6 1 4

= = + +

1 1

10 10=


( ) ( ) ( ) ( )1 1 1 1

P ..........2.5 5.8 8.11 3 1 3 2 6 4

kk

k k k= + + + + =

+ +

For 1n k= +

( ) ( )( ) ( ) ( )1 1 1 1 1

P 1 ..........2.5 5.8 8.11 3 1 3 2 3 1 1 3 1 2

kk k k k

+ = + + + + + + + + +

( ) ( ) ( )1 1

6 1 4 3 1 1 3 1 2

k

k k k

+= ++ + + + +



( ) ( )1 1

P 13 2 2 3 5

kk

k k + = + + +

= ( ) ( )

21 3 5 2

3 2 2 3 5

k k

k k

+ + + +

= ( )

( )( )( )1 3 21

3 2 2 3 5

k k

k k

+ + + +

( ) 1P 16 10

kk

k

++ =+


11. Let ( ) ( ) ( )( )

( ) ( )31 1 1 1

P ..........1.2.3 2.3.4 3.4.5 1 2 4 1 2

n nn

n n n n n

+= + + + + =

+ + + +

For 1n = ( ) ( ) ( )( )

( )( )1 1 31

P 11 1 1 1 2 4 1 1 1 2

+= =

+ + + +

1 1

6 6=


( ) ( )( )( )

( ) ( )31 1 1 1

P ..........1.2.3 2.3.4 3.4.5 1 2 4 1 2

k kk

k k k k k

+= + + + + =

+ + + + .(i)

For 1n k= +

R.H.S. = ( ) ( )( ) ( )

1 4

4 2 3

k k

k k

+ ++ +

And L.H.S. = ( )

( )( ) ( )( )( )3 1

4 2 3 1 2 3

k k

k k k k k

++

+ + + + + [Using eq. (i)]

= ( ) ( )21 3 1

1 2 4 3

k k

k k k

+ + + + + =

( ) ( ) ( )3 21 6 9 4

1 2 4 3

k k k

k k k

+ + + + + +

= ( ) ( )( ) ( )

( )

21 41

1 2 4 3

k k

k k k

+ +

+ + + =

( ) ( )( ) ( )

1 4

4 2 3

k k

k k

+ ++ +


12. Let ( ) ( )2 1 1P ..........1

n

na r

n a ar ar arr

= + + + + =

For 1n = ( ) ( )1

1 11

P 11

a rar

r

= =

a a=




( ) ( )2 1 1P ..........1

k

ka r

k a ar ar arr

= + + + + =

.(i)

For 1n k= +

R.H.S. = ( )1 1

1

ka r

r

+

L.H.S. =

( )1 11

k

ka r

arr

+ +

[Using eq. (i)]

L.H.S. = 1 1

kkar a ar

r r +

=

11

1 1k aar

r r +

= 1 1

k r aarr r

= 1

1 1

kar a

r r

+

= 1

1

kar a

r

+

= ( )1 1

1

ka r

r

+


13. Let ( ) ( )223 5 7 2 1

P 1 1 1 .......... 1 11 4 9

nn n

n

+ = + + + + = +

For 1n = ( ) ( ) ( )222 1 1

P 1 1 1 11

+= + = + 4 4=


( ) ( )223 5 7 2 1

P 1 1 1 .......... 1 11 4 9

kk k

k

+ = + + + + = +

.(i)

For 1n k= +

R.H.S. = ( )22k + L.H.S. = ( )( )

2

2

2 31 1

1

kk

k

++ + +

[Using eq. (i)]

L.H.S. = ( ) ( )( )

22

2

1 2 31

1

k kk

k

+ + ++

+ = ( )22 4 4 2k k k+ + = +


14. Let ( ) ( )1 1 1 1P 1 1 1 .......... 1 11 2 3

n nn

= + + + + = +

For 1n = ( ) 1P 1 1 1 11

= + = +

2 2=

( )P 1 is true.



Now, let ( )P n be true for n k=

( ) ( )1 1 1 1P 1 1 1 .......... 1 11 2 3

k kk

= + + + + = +

For 1n k= + R.H.S. = 2k +

L.H.S. = ( ) 11 11

kk

+ + + [Using eq. (i)]

L.H.S. = ( ) 1 111

kk

k

+ + + + = ( )2k +


15. Let ( ) ( ) ( )( )22 2 2 2 1 2 1P 1 3 5 .......... 2 13

n n nn n

+= + + + + =

For 1n = ( ) ( ) ( )( )2 1 2 1 1 2 1 1P 1 2 1 13

+= = 1 1=


( ) ( ) ( )( )22 2 2 2 1 2 1P 1 3 5 .......... 2 13

k k kk k

+= + + + + = .(i)

For 1n k= + R.H.S. = ( )( )( )1 2 1 2 33

k k k+ + +

L.H.S. = ( )( ) ( )22 1 2 1 2 1

3

k k kk

++ + [Using eq. (i)]

= ( ) ( ) ( )2 12 1 2 13

k kk k

+ + + +

= ( )

2 6 32 1

3

k k kk

+ ++

= ( ) ( )22 1 2 5 3

3

k k k+ + +

= ( ) ( )( )2 1 1 2 3

3

k k k+ + + =

( )( )( )1 2 1 2 33

k k k+ + +


16. Let ( ) ( )( ) ( )1 1 1 1

P ..........1.4 4.7 7.10 3 2 3 1 3 1

nn

n n n= + + + + =

+ +

For 1n = ( ) ( )( ) ( )1 1

P 13 1 2 3 1 1 3 1 1

= = + +

1 1

4 4=




( ) ( )( ) ( )1 1 1 1

P ..........1.4 4.7 7.10 3 2 3 1 3 1

kk

k k k= + + + + =

+ + .(i)

For 1n k= + R.H.S. = 13 1

k

k

++

L.H.S. = ( )( )1

3 1 3 1 3 4

k

k k k+

+ + +

L.H.S. = 1 1

3 1 3 4k

k k + + +

= 21 3 4 1

3 1 3 4

k k

k k

+ + + +

= ( ) ( )3 1 11 1

3 1 3 4 3 4

k k k

k k k

+ + += + + +


17. Let ( ) ( ) ( ) ( )1 1 1 1

P ..........3.5 5.7 7.9 2 1 2 3 3 2 3

nn

n n n= + + + + =

+ + +

For 1n = ( ) ( ) ( ) ( )1 1

P 12 1 1 2 1 3 3 2 1 3

= = + + +

1 1

15 15=


( ) ( ) ( ) ( )1 1 1 1

P ..........3.5 5.7 7.9 2 1 2 3 3 2 3

kk

k k k= + + + + =

+ + + .(i)

For 1n k= + R.H.S. = ( )1

3 2 5

k

k

++

L.H.S. = ( ) ( ) ( )1

3 2 3 2 3 2 5

k

k k k+

+ + +

L.H.S. = ( )1 1

2 3 3 2 5

k

k k + + +

= ( ) ( )

21 2 5 3

2 3 3 2 5

k k

k k

+ + + +

= ( )

( )( )( )1 2 31

2 3 3 2 5

k k

k k

+ + + +

= ( )1

3 2 5

k

k

++


18. Let ( ) ( )21P 1 2 3 .......... 2 18

n n n= + + + + < +

For 1n = ( ) ( )21P 1 1 2 1 18

= < + 918

<




( ) ( )21P 1 2 3 .......... 2 18

k k k= + + + + < + .(i)

For 1n k= + , ( ) ( ) ( ) ( )21P 1 1 2 3 .......... 1 2 1 18

k k k k k+ = + + + + + + < + + +

Now, adding ( )1k + on both sides of eq. (i), we have

( ) ( ) ( )211 2 3 .......... 1 2 1 18

k k k k+ + + + + + < + + +

( ) ( )211 2 3 .......... 1 4 4 1 8 88

k k k k k+ + + + + + < + + + +

( ) ( )211 2 3 .......... 1 4 12 98

k k k k+ + + + + + < + +

( ) ( )211 2 3 .......... 1 2 38

k k k+ + + + + + < +

( )( ) ( )21 2 1 2 3

2 8

k kk

+ +< + ( )( ) 24 1 2 4 9 6k k k k+ + < + +

( )2 24 3 2 4 9 6k k k k+ + < + + 2 24 12 8 4 9 6k k k k+ + < + + 8 < 9


19. Let ( ) ( )( )P 1 5n n n n= + + is a multiple of 3. For 1,n = ( )P 1 = 1 (1 + 1) (1 + 5) is a multiple of 3 = 12 is a multiple of 3 P (1) is true. Let ( )P n be true for n k= , ( ) ( ) ( )P 1 5k k k k= + + is a multiple of 3. ( ) ( )1 5 3k k k + + = 3 26 5 3k k k + + = 3 23 6 5k k k= .(i) For 1n k= + , ( ) ( ) ( ) ( )P 1 1 2 6k k k k+ = + + + is a multiple of 3 Now, ( ) ( ) ( )1 2 6k k k+ + + = 3 29 20 12k k k+ + + = 2 23 6 5 9 20 12k k k k + + + [Using (i)] = 23 3 15 12k k + + + = ( )23 5 4k k + + + = ( ) ( ) ( )1 2 6k k k+ + + is a multiple of 3 ( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

20. Let ( ) 2 1P 10 1nn = + is divisible by 11. For 1,n = ( ) 2 1 1P 1 10 1 = + is divisible by 11 = 11 is divisible by 11 P (1) is true.



Let ( )P n be true for n k= , ( ) 2 1P 10 1kk = + is divisible by 11 = 2 110 1 11k + = 2 110 11 1k = .(i) For 1,n k= + ( ) ( )2 1 1P 1 10 1kk + + = + is divisible by 11 ( ) 2 1P 1 10 1kk ++ = + is divisible by 11 Now, 2 1 2 1 210 1 10 .10 1k k+ + = + = ( ) 211 1 .10 1 1100 100 1 + = + = ( )11 100 9 ( )2 1 110 1k+ + is divisible by 11 ( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

21. Let ( ) 2 2P n nn x y= is divisible by ( )x y+ For 1,n = ( ) 2 1 2 1P 1 x y = is divisible by ( )x y+ = ( )( )x y x y+ is divisible by ( )x y+ P (1) is true. Let ( )P n be true for n k= , ( ) 2 2P k kk x y= is divisible by ( )x y+ = ( )2 2k kx y x y = + ( )2 2k kx y x y = + .(i) For 1,n k= + ( ) ( ) ( )2 1 2 1P 1 k kk x y+ ++ = is divisible by ( )x y+ Now, 2 2 2 2 2 2 2 2 2 2 2 2k k k k k kx y x x y x y y+ + = + = 2 2 2 2 2 2 2 2. .k k k kx x x y x y y y +

= ( ) ( )2 2 2 2 2 2k k kx x y y x y + = ( ) ( )2 2 2 2.kx x y y x y + + [From eq. (i)] = ( ) ( )2 2kx y x x y y + +

( ) ( )2 1 2 1k kx y+ + is divisible by ( )x y+ ( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

22. Let ( ) 2 2P 3 8 9nn n+= is divisible by 8. For 1,n = ( ) 2 1 2P 1 3 8 1 9 += is divisible by 8 = 64 is divisible by 8 P (1) is true. Let ( )P n be true for n k= , ( ) 2 2P 3 8 9kk k+= is divisible by 8 = 2 23 8 9 8k k + = 2 23 8 8 9k k+ = + + .(i) For 1,n k= + ( ) ( ) ( )2 1 2P 1 3 8 1 9kk k+ ++ = + is divisible by 8 ( ) 2 2 2P 1 3 .3 8 8 9kk k++ = is divisible by 8 Now, 2 23 .9 8 17k k+ = ( )8 8 9 .9 8 17k k + + [From eq. (i)] = 72 72 81 8 17k k + + = 72 64 64k + + = ( )8 9 8 8k + +



( ) ( )2 1 23 8 1 9k k+ + + is divisible by 8 ( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

23. Let ( )P 41 14n nn = is a multiple of 27. For 1,n = ( ) 1 1P 1 41 14= is a multiple of 27 = 27 is a multiple of 27 P (1) is true. Let ( )P n be true for n k= , ( )P 41 14k kk = is a multiple of 27 = 41 14 27k k = ..(i) For 1,n k= + ( ) 1 1P 1 41 14k kk + ++ = is a multiple of 27 Now, 1 141 14k k+ + = 1 .14 141 41 .14 41 14k k k k+ + +

= ( ) ( )41 41 14 14 41 14k k k + = 41 27 14 27k + [From eq. (i)] = ( )27 41 14k + 1 141 14k k+ + is a multiple of 27 ( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

24. Let ( ) ( ) ( )2P 2 7 3n n n= + < + For 1,n =

( ) ( ) ( )2P 1 2 1 7 1 3= + < + 9 < 16

P (1) is true. Let ( )P n be true for n k=

( ) ( ) ( )2P 2 7 3k k k= + < + .(i) For 1n k= +

( ) ( ) ( )2P 1 2 1 7 1 3k k k+ = + + < + + = ( ) ( )22 1 7 4k k+ + < + Now, adding 2 on both sides in eq. (i),

( ) ( )22 7 2 3 2k k+ + < + + ( ) 22 1 7 9 6 2k k k+ + < + + + ( ) 22 1 7 6 11k k k+ + < + + ( ) 22 1 7 6 11k k k+ + < + +



Also ( ) ( )222 1 7 6 11 4k k k k+ + < + + < + ( ) ( )22 1 7 4k k+ + < + ( )P 1k + is true. Therefore, ( )P k is true. ( )P 1k + is true. Hence by Principle of Mathematical Induction, ( )P n is true for all n N.

Documents

11 Mathematics Ncert Ch04 Principle of Mathematical Induction 4.1 Solutions