CLASS XI CHAPTER 4 PRINCIPLE OF MATHEMATICAL INDUCTION RR ACADEMY MEERUT

Chapter 4: Principle of Mathematical Induction

Exercise 4.1

Prove the following through principle of mathematical induction for all values of n, where n is a natural number.

1) 1+3+32+….+3n–1=(3n–1)2

Sol:

The given statement is:

P(n) : 1+3+32+….+3n–1=(3n–1)2

Now, for n = 1

P(1) = (31–1)2 = (3–1)2= 22= 1

Thus, the P(n) is true for n=1

Let,[2(k+1)+7]=[(2k+7)+2][2(k+1)+7]=[(2k+7)+2][2(k+1)+7]=[(2k+7)+2]

P(k) be true, where k is a positive integer.

1+3+32+….+3k–1=(3k–1)2 . . . . . . . . . . (1)

Now, we will prove that P(k+1) is also true:

P(k + 1):

=1+3+32+….+3k–1+3(k+1)–1

=(1+3+32+….+3k–1)+3k        [Using equation (1)]

=(3k–1)2+3k

=(3k–1)+2.3k2

=(1+2)3k–12

=3.3k–12

=3k+1–12

Thus, whenever P(k) proves to be true, P(k+1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

2: 13+23+33+……+n3 = (n(n+1)2)2

Sol:

The given statement is:

P(n): 13+23+33+……+n3 = (n(n+1)2)2

Now, for n = 1

P(1): 13=1=(1(1+1)2)2= (1×22)2= 12 = 1

Thus, the P(n) is true for n = 1

Let, P(k) be true, where ‘k’ is a positive integer.

13+23+33+……+k3=(k(k+1)2)2 . . . . . . . . . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true.

P(k + 1):

=13+23+33+……+k3+(k+1)3

=(13+23+33+……+k3)+(k+1)3

=(k(k+1)2)2+(k+1)3    [From equation (1)]

=k2(k+1)24+(k+1)3

=k2(k+1)2+4(k+1)34

=(k+1)2{k2+4(k+1)}4

=(k+1)2{k2+4k+4}4

=(k+1)2(k+2)24

=(k+1)2(k+1+1)24

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

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

3: 1+11+2+11+2+3+….+11+2+3+…+n=2nn+1

Sol:

The given statement is:

P(n): 1+11+2+11+2+3+….+11+2+3+…+n=2nn+1

Now, for n = 1

P(1): 1=2×11+1=22=1

Thus, the P(n) is true for n=1

Let, P(k) be true, where k is a positive integer.

1+11+2+11+2+3+….+11+2+3+…+k=2kk+1 . . . . . . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true.

=1+11+2+11+2+3+….+11+2+3+…+k+11+2+3+…+k+(k+1)

=[1+11+2+11+2+3+….+11+2+3+…+k]+11+2+3+…+k+(k+1)

=2kk+1+11+2+3+…+k+(k+1)      [From equation (1)]

= 2kk+1+1((k+1)(k+1+1)2)            (1+2+3+…+n=n(n+1)2)

= 2kk+1+2(k+1)(k+2)

= 2(k+1)(k+1k+2)

= 2(k+1)(k2+2k+1k+2)

= 2(k+1)((k+1)2k+2)

= 2(k+1)(k+2)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

4: 1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4

Sol:

The given statement is:

P(n): 1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4

Now, for n = 1

P(1): 1.2.3=6= 1(1+1)(1+2)(1+3)4= 1.2.3.44= 6

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

1.2.3+2.3.4+…+k(k+1)(k+2)=k(k+1)(k+2)(k+3)4  . . . . . . . . . . (1)

Now, we will prove P(k + 1) is true.

=1.2.3+2.3.4+…+k(k+1)(k+2)+(k+1)(k+2)(k+3)

=[1.2.3+2.3.4+…+k(k+1)(k+2)]+n(k+1)(k+2)(k+3)     [By using equation (1)]

=k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3)

Now, by using equation (1) :

=(k+1)(k+2)(k+3)(k4+1)

=(k+1)(k+2)(k+3)(k+4)4

=(k+1)(k+1+1)(k+1+2)(k+1+3)4

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

5: 1.3+2.32+3.33+…+n.3n=(2n−1)3n+1+34

Sol:

The given statement is:

P(n): 1.3+2.32+3.33+…+n.3n=(2n−1)3n+1+34

Now, for n = 1:

=(2.1–1)31+1+34= 32+34= 124= 3

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

1.3 + 2. 3^{2}+ 3.3^{3} +…+ k. 3^{k} = \frac{(2k-1)3^{k+1}\, +\, 3}{4} . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=1.3+2.32+3.33+…+k.3k+(k+1).3k+1

=(2k−1)3k+1+34+(k+1)3k+1     [By using equation (1)]

=(2k−1)3k+1+3+4(k+1)3k+14

= 3k+1{2k–1+4(k+1)}+34

=3k+1{6k+3}+34

=3k+1.3{2k+1}+34

= 3(k+1)+1{2k+1}+34

={2(k+1)−1}3(k+1)+1+34

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

6: 1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]

Sol:

The given statement is:

P(n): 1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]

Now, for n = 1:

=1(1+1)(1+2)3= 1.2.33= 2

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

1.2+2.3+3.4+…+k.(k+1)=[k(k+1)(k+2)3]  . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=[1.2+2.3+3.4+…+k.(k+1)]+(k+1)(k+2)

=k(k+1)(k+2)3+(k+1)(k+2)

Now, by using equation (1):

=(k+1)(k+2)(k3+1)

=(k+1)(k+2)(k+3)3

=(k+1)(k+1+1)(k+1+2)3

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

7: 1.3+3.5+5.7+…+(2n–1)(2n+1)=n(4n2+6n–1)3

Sol:

The given statement is:

1.3+3.5+5.7+…+(2n–1)(2n+1)=n(4n2+6n–1)3

Now, for n = 1:

1(4.12+6.1–1)3= 4+6–13= 93 = 3

Thus, the P(n) is true for n = 1

Let, P(k) be true, where k is a positive integer.

1.3+3.5+5.7+…+(2k–1)(2k+1)=k(4k2+6k–1)3 . . . . . . . . .  (1)

Now we will prove P(k + 1) is also true:

=1.3+3.5+5.7+…+(2k–1)(2k+1)+{2(k+1)–1}{2(k+1)+1}

=k(4k2+6k–1)3+(2k+2–1)(2k+2+1)

Now, by using equation (1):

=k(4k2+6k–1)3+(2k+1)(2k+3)

=k(4k2+6k–1)3+(4k2+8k+3)

=k(4k2+6k–1)+3(4k2+8k+3)3

=4k3+6k2–k+12k2+24k+93

=4k3+18k2+23k+93

=4k3+14k2+9k+4k2+14k+93

=k(4k2+14k+9)+1(4k2+14k+9)3

=(k+1)(4k2+14k+9)3

=(k+1)(4k2+8k+4+6k+6–1)3

=(k+1){4(k2+2k+1)+6(k+1)–1}3

=(k+1){4(k+1)2+6(k+1)–1}3

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

8: 1.2+2.22+3.22+…+n.2n=(n–1)2n+1+2

Sol:

The given statement is:

P(n): 1.2+2.22+3.22+…+n.2n=(n–1)2n+1+2

Now, for n = 1:

=(1–1)21+1+2 = 0 + 2= 2

Thus, the P(n) is true for n=1.

Let P(k) be true, where k is a positive integer:

1.2+2.22+3.22+…+k.2k=(k–1)2k+1+2 . . . . . . . . (1)

Now we will prove P(k + 1) is also true:

=[1.2+2.22+3.22+…+k.2k]+(k+1).2k+1

=(k–1)2k+1+2+(k+1).2k+1

=2k+1{(k–1)+(k+1)}+2

=2k+1.2k+2

=k.2(k+1)+1+2

={(k+1)–1}2(k+1)+1+2

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

9: 12+14+18+…+12n=1–12n

Sol:

The given statement is:

12+14+18+…+12n=1–12n

Now, for n = 1:

=1–121= 12

Thus, the P (n) is true for n = 1.

Let, P (k) be true, where k is a positive integer:

12+14+18+…+12k=1–12k . . . . . . (1)

Now, we will prove P (k + 1) is also true:

=(12+14+18+…+12k)+12k+1

=(1–12k)+12k+1

Now, by using equation (1):

=1–12k+121.2k

=1–12k(1–12)

=1–12k(12)

=1–12k+1

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 10: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

12⋅5+15⋅8+18⋅11+……+1(3n−1)(3n+2)=n(6n+4)

Solution:

The given Statement is:
Q(n): 12⋅5+15⋅8+18⋅11+……+1(3n−1)(3n+2)=n(6n+4)

Now, for n = 1

Q(1)=12⋅5=1(6(1)+4)=12⋅5

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

= 12⋅5+15⋅8+18⋅11+……+1(3p−1)(3p+2)=p(6p+4) . . . . . . . . . . . . . . (1)

Now, we have to  prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

= 12⋅5+15⋅8+18⋅11+……+1(3p−1)(3p+2)+1[3(p+1)−1][3(p+1)+2]

Now, by using equation (1):

=p(6p+4)+1[3p+3–1][3p+3+2]

=p(6p+4)+1(3p+2)(3p+3+2)

=p2(3p+2)+1[3p+3–1][3p+3+2]

=p2(3p+2)+1[3p+3–1][3p+3+2]

=12(3p+2)[p2+1(3p+5)]

=12(3p+2)[p(3p+5)+2(3p+5)]

=12(3p+2)[3p2+5p+2(3p+5)]

=12(3p+2)[(3p+2)(p+1)(3p+5)]

=p+16p+10

=p+16(p+1)+4

Thus, whenever Q (p) proves to be true, Q (p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

11: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

11⋅2⋅3+12⋅3⋅4+13⋅4⋅5+…….+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

Solution:

The given statement is:

Q(n): 11⋅2⋅3+12⋅3⋅4+13⋅4⋅5+…….+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

Now, for n = 1:

Q(1)=11⋅2⋅3=(1)((1)+3)4((1)+1)((1)+2)=11⋅2⋅3

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number:

Q (p):

=11⋅2⋅3+12⋅3⋅4+13⋅⋅4⋅5+…….+1p(p+1)(p+2)=p(p+3)4(p+1)(p+2). . . . . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q (p + 1):

=p(p+3)4(p+1)(p+2)+1(p+1)(p+2)(p+3)

Now, using equation (1):

=1(p+1)(p+2)[p(p+3)4+1(p+3)]

=1(p+1)(p+2)[p(p+3)2+44(p+3)]

=1(p+1)(p+2)[p(p2+6p+9)+44(p+3)]

=1(p+1)(p+2)[p(p2+6p+9)+44(p+3)]

=1(p+1)(p+2)[p3+6p2+9p+44(p+3)]

=1(p+1)(p+2)[p(p2+2p+1)+4(p2+2p+1)4(p+3)]

=1(p+1)(p+2)[p(p+1)2+4(p+1)24(p+3)]

=1(p+1)(p+2)[(p+1)2(p+4)4(p+3)]

=(p+1)[(p+1)+3]4[(p+1)+1][(p+1)+2]

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

12: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

a+ar+ar2+…….+arn−1=a(rn−1)r−1

Solution:

The given statement is:

Q(n): a+ar+ar2+…….+arn−1=a(rn−1)r−1

Now, for n = 1

Q(1)=a=a(r(1)−1)r−1=a

Thus, Q (1) proves to be true.

Let’s assume Q (p) is true, where p is a natural number.

Q (p)  = a+ar+ar2+…….+arp−1=a(rp−1)r−1 . . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q(p+1) = a+ar+ar2+…….+arp−1+ar(p+1)−1

Now, using Equation (1):

=a(rp–1)r–1+arp

=a(rp–1)+arp(r–1)r–1

=a(rp–1)+arp(r–1)–arpr–1

=arp–a+arp+1–arpr–1

=arp+1–ar–1

=a(rp+1–1)r–1

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

13: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

(1+31)(1+54)(1+79)…….(1+(2n+1)n2)=(n+1)2

Solution:

The given statement is:

Q(n): (1+31)(1+54)(1+79)…….(1+(2n+1)n2)=(n+1)2

Now, for n = 1:

Q(1)=(1+31)=4 ⇒(n+1)2=(1+1)2=4

Thus, Q (1) proves to be true.

Let’s assume Q (p) is true, where p is a natural number.

Q (p):

=(1+31)(1+54)(1+79)…….(1+(2p+1)p2)=(p+1)2 . . . . . . . . . . . (1)

Now we have to prove that Q(p+1) is also true.

Since Q (p) is true, we have:

Q(p+1):

=(1+31)(1+54)(1+79)…….(1+(2p+1)p2)(1+(2(p+1)+1(p+1)2)

Now, using equation (1):

=(p+1)2(1+(2(p+1)+1(p+1)2)

=(p+1)2((p+1)2+2(p+1)+1(p+1)2)

⇒ (p+1)2+2(p+1)+1=[(p+1)+1]2
Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

14: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

(1+11)(1+12)(1+13)…….(1+1n)=(n+1)

Solution:

The given statement is:

Q(n): (1+11)(1+12)(1+13)…….(1+1n)=(n+1)

Now, for n = 1:

Q(1)=(1+11)=2⇒(n+1)=1+1=2

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

=(1+11)(1+12)(1+13)…….(1+1p)=(p+1). . . . .(1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

=(1+11)(1+12)(1+13)…….(1+1p)(1+1p+1)

Now, using Equation (1):

=(p+1)(1+1p+1)=(p+1)((p+1)+1p+1)

= (p + 1) + 1

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

15: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

12+32+52+.…..+(2n−1)2=n(2n−1)(2n+1)3

Solution:

The given statement is:

Q(n): 12+32+52+.…..+(2n−1)2=n(2n−1)(2n+1)3

Now, for n = 1:

Q(1)=12=1and(1)(2(1)−1)(2(1)+1)3=33=1

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q (p):

=12+32+52+.…..+(2p−1)2=p(2p−1)(2p+1)3 . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q(p) is true, we have:

Q (p+1):

=12+22+32+……+(2k–1)2]+(2(p+1)–1)2

=p(2p–1)(2p+1)3+(2p+2−1)2

=p(2p–1)(2p+1)3+(2p+1)2

=p(2p–1)(2p+1)+3(2p+1)23

=(2p+1)(p(2p–1)+3(2p+1))3

=(2p+1)(2p2–p+6p+3)3

=(2p+1)(2p2+5p+3)3

=(2p+1)(2p2+2p+3p+3)3

=(2p+1)(2p(p+1)+3(p+1))3

=(2p+1)(p+1)(2p+3)3

=(p+1)[2(p+1)–1][2(p+1)+1]3

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

16: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

11⋅4+14⋅7+17⋅10+……+1(3n−2)(3n+1=n3n+1

Solution:

The given statement is:

Q(n): 11⋅4+14⋅7+17⋅10+……+1(3n−2)(3n+1=n3n+1

Now, for n = 1:

Q(1)=11.4=1(3(1)−2)(3(1)+1)=11.4

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

=11⋅4+14⋅7+17⋅10+……+1(3p–2)(3p+1)=p3p+1 . . . . . . . . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

=11⋅4+14⋅7+17⋅10+…+1(3p–2)(3p+1)+1[3(p+1)–2][3(p+1)+1]

Now, using Equation (1):

=1(3p+1)+1(3p+1)(3p+4)]

=1(3p+1)[p+1(3p+4)]

=1(3p+1)[p(3p+4)+1(3p+4)]

=1(3p+1)[3p2+4p+1(3p+4)]

=1(3p+1)[3p2+3p+p+1(3p+4)]

=1(3p+1)[(3p+1)(p+1)(3p+4)]

=(p+1)3(p+1)+1

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

17: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

13⋅5+15⋅7+17⋅9+…….+1(2n+1)(2n+3)=n3(2n+3)

Solution:

The given statement is:

Q(n): 13⋅5+15⋅7+17⋅9+…….+1(2n+1)(2n+3)=n3(2n+3)

Now, for n = 1:

Q(1)=13⋅5=13(2(1)+3)=13⋅5

Thus, Q(1) proves to be true:

Let us assume that Q(p) is true, where p is a natural number.

Therefore, Q(p):

=13⋅5+15⋅7+17⋅9+…….+1(2p+1)(2p+3)=p3(2p+3) . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q(p) is true, we have:

Q(p+1):

=13⋅5+15⋅7+17⋅9+…….+1(2p+1)(2p+3)+1[2(p+1)+1][2(p+1)+3]

Now, Using Equation (1):

=p3(2p+3)+1(2p+3)(2p+5)

=13(2p+3)[p3+1(2p+5)]

=13(2p+3)[p(2p+5)+3(2p+5)]

=13(2p+3)[2p2+5p+3(2p+5)]

=13(2p+3)[2p2+2p+3p+3(2p+5)]

=13(2p+3)[2p(p+1)+3(p+1)(2p+5)]

=13(2p+3)[(2p+3)(p+1)(2p+5)]

=p+13[2(p+1)+3]

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

18: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

1+2+3+……+n<18(2n+1)2

Solution:

The given statement is:

Q(n): 1+2+3+……+n<18(2n+1)2

Now, for n = 1:

Q(1)=1=18(2(1)+1)2=98⇒1<98

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p) = 1+2+3+……+p<18(2p+1)2 . . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Now, Using Equation (1):

=(1+2+3+……+p)+(p+1)<18(2p+1)2+(p+1)

<18[(2p+1)2+8(p+1)]

<18[4p2+4p+1+8p+8]

<18[4p2+12p+9]

<18[(2p+3)2]

<18[(2(p+1)+1)2]

Thus, (1+2+3+……+p)+(p+1)<18(2p+1)2+(p+1)

Thus, whenever Q (p) proves to be true, Q (p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q (n) is true for all the natural numbers.

19: n (n + 1)(n + 5) is a multiple of 3.

Sol:

The given statement is:

P(n): n (n + 1) (n + 5) is a multiple of 3

Now, for = 1:

= 1(1 + 1)(1 + 5) = 12

Thus, the P(n) is true for n = 1.

Let, P(k) be true, where k is a positive integer.

k(k + 1)(k + 5) is a multiple of 3

Therefore, k(k + 1)(k + 5) = 3m, where m ∈ N . . . . . . . . (1) 

Now, we will prove P(k + 1) is also true.

=(k+1){(k+1)+1}{(k+1)+5}

= (k+1)(k+2){(k+5)+1}

= (k+1)(k+2)(k+5)+(k+1)(k+2)

= {k(k+1)(k+5)+2(k+1)(k+5)}+(k+1)(k+2)

= 3m+(k+1){2(k+5)+(k+2)}

= 3m+(k+1){2k+10+k+2}

= 3m+(k+1){3k+12}

= 3m+3(k+1)(k+4)

= 3{m+(k+1)(k+4)}

= 3×q, where q = {m+(k+1)(k+4)} ∈ N.

Therefore, (k+1){(k+1)+1}{(k+1)+5} is a multiple of 3.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

20: 102n–1+1 is divisible by 11.

Sol:

The given statement is:

102n–1+1 is divisible by 11

Now, for n = 1

=102.1–1+1 = 11

Thus, the P(n) is true for n = 1.

Then, P(k) is also true, where k is a positive integer.

102k–1+1 is divisible by 11.

102k–1+1 = 11m, where m ∈ N  . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=102(k+1)−1+1

=102k+2–1+1

=102k+1+1

=102(102k–1+1–1)+1

=102(102k–1+1)–102+1

=102.11m–100+1          [using equation (1)]

=100×11m–99

=11(100m–9)

= 11 r, where r = (100m–9) is some natural number.

Therefore,  102(k+1)−1+1 is divisible by 11.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

21. x2n–y2n is divisible by x + y.

Sol:

The given statement is:

P(n): x2n–y2n is divisible by x + y.

Now, for n = 1

= x2×1–y2×1= x2–y2= (x + y) (x – y)

Therefore, it is divisible by (x + y).

Thus, the P(n) is true for n=1.

Let, P(k) is also true, where k is a positive integer.

x2k–y2k is divisible by (x + y).

Let x2k–y2k = m (x + y), where m ∈ N . . . . (1)

Now, we will prove P(k + 1) is also true:

=x2(k+1)–y2(k+1)

=x2k.x2–y2k.y2

=x2(x2k–y2k+y2k)–y2k.y2

=x2{m(x+y)+y2k}–y2k.y2             [using equation (1)]

=m(x+y)x2+y2k.x2–y2k.y2

=m(x+y)x2+y2k(x2–y2)

=m(x+y)x2+y2k(x+y)(x–y)

=(x+y){mx2+y2k(x–y)}, which is the factor of (x + y).

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

22: 32n+2–8n–9 is divisible by 8.

Sol:

The given statement is:

P(n): 32n+2–8n–9 is divisible by 8.

Now, for n = 1:

=32.1+2–8.1–9= 34–17= 64

Therefore, it is divisible by 8.

Thus, the P(n) is true for n = 1.

Let, P(k) be true, where k is a positive integer.

32k+2–8k–9 is divisible by 8.

32k+2–8k–9=8m, where m ∈ N. . . . . . (1)

Now, we will prove P(k + 1) is also true:

=32(k+1)+2–8(k+1)–9

=32k+2.32–8k–8–9

=32(32k+2–8k–9+8k+9)–8k–17

=32(32k+2–8k–9)+32(8k+9)–8k–17

=9.8m+9(8k+9)–8k–17

=9.8m+72k+81–8k–17

=9.8m+64k+64

=8(9m+8k+8)

= 8r, where r = (9m+8k+8) is a natural number.

Therefore, 32(k+1)+2–8(k+1)–9 is divisible by 8.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

23: 41n–14n is a multiple of 27.

Sol:

The given statement is:

P(n): 41n–14n is a multiple of 27

Now, for n = 1:

=411–141= 27, which is the multiple of 27

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

41k–14k is a multiple of 27.

41k–14k = 27m , where m ∈ N . . . . . . . . (1)

Now, we will prove P(k + 1) also is true:

=41k+1–14k+1

=41k.41–14k.14

=41(41k–14k+14k)–14k.14

=41(41k–14k)+41.14k–14k.14

=41.27m+14k(41–14)

=41.27m+27.14k

=27(41m+14k)

=27×r, where r = (41m+14k) is a natural number

Therefore, 41k+1–14k+1 is a multiple of 27.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

24: (2n+7)<(n+3)2

Sol:

The given statement is:

P(n): (2n+7)<(n+3)2

Now, for n = 1:

=(2.1+7)<(1+3)2= (9)<(4)2= 9<16

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

(2k+7)<(k+3)2 . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=[2(k+1)+7]=[(2k+7)+2]

=[2(k+1)+7] = (2k+7)+2 < (k+3)2+2     [using equation (1)]

=2(k+1)+7<k2+6k+9+2

=2(k+1)+7<k2+6k+11

Now,k2+6k+11<k2+8k+16

2(k+1)+7<(k+4)4

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.