Class 11 Maths Ncert Solutions Ex 4.1

Class 11 Maths Ncert Solutions Chapter 4 Ex 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+.+3n1=(3n1)2

 

Sol:

The given statement is:

P(n) : 1+3+32+.+3n1=(3n1)2

Now, for n = 1

P(1) = (311)2 = (31)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+.+3k1=(3k1)2 . . . . . . . . . . (1)

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

P(k + 1):

=1+3+32+.+3k1+3(k+1)1

=(1+3+32+.+3k1)+3k        [Using equation (1)]

=(3k1)2+3k

=(3k1)+2.3k2

=(1+2)3k12

=3.3k12

=3k+112

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=(2n1)3n+1+34

 

Sol:

The given statement is:

P(n): 1.3+2.32+3.33++n.3n=(2n1)3n+1+34

Now, for n = 1:

=(2.11)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

=(2k1)3k+1+34+(k+1)3k+1     [By using equation (1)]

=(2k1)3k+1+3+4(k+1)3k+14

= 3k+1{2k1+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++(2n1)(2n+1)=n(4n2+6n1)3

 

Sol:

The given statement is:

1.3+3.5+5.7++(2n1)(2n+1)=n(4n2+6n1)3

Now, for n = 1:

1(4.12+6.11)3= 4+613= 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++(2k1)(2k+1)=k(4k2+6k1)3 . . . . . . . . .  (1)

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

=1.3+3.5+5.7++(2k1)(2k+1)+{2(k+1)1}{2(k+1)+1}

=k(4k2+6k1)3+(2k+21)(2k+2+1)

Now, by using equation (1):

=k(4k2+6k1)3+(2k+1)(2k+3)

=k(4k2+6k1)3+(4k2+8k+3)

=k(4k2+6k1)+3(4k2+8k+3)3

=4k3+6k2k+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+61)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=(n1)2n+1+2

 

Sol:

The given statement is:

P(n): 1.2+2.22+3.22++n.2n=(n1)2n+1+2

Now, for n = 1:

=(11)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=(k1)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

=(k1)2k+1+2+(k+1).2k+1

=2k+1{(k1)+(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=112n

 

Sol:

The given statement is:

12+14+18++12n=112n

Now, for n = 1:

=1121= 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=112k . . . . . . (1)

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

=(12+14+18++12k)+12k+1

=(112k)+12k+1

Now, by using equation (1):

=112k+121.2k

=112k(112)

=112k(12)

=112k+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.

125+158+1811++1(3n1)(3n+2)=n(6n+4)

 

Solution:

The given Statement is:
Q(n): 125+158+1811++1(3n1)(3n+2)=n(6n+4)

Now, for n = 1

Q(1)=125=1(6(1)+4)=125

Thus, Q(1) proves to be true.

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

Q(p):

= 125+158+1811++1(3p1)(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):

= 125+158+1811++1(3p1)(3p+2)+1[3(p+1)1][3(p+1)+2]

Now, by using equation (1):

=p(6p+4)+1[3p+31][3p+3+2]

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

=p2(3p+2)+1[3p+31][3p+3+2]

=p2(3p+2)+1[3p+31][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.

1123+1234+1345+.+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

 

Solution:

The given statement is:

Q(n): 1123+1234+1345+.+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

Now, for n = 1:

Q(1)=1123=(1)((1)+3)4((1)+1)((1)+2)=1123

Thus, Q(1) proves to be true.

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

Q (p):

=1123+1234+1345+.+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+.+arn1=a(rn1)r1

 

Solution:

The given statement is:

Q(n): a+ar+ar2+.+arn1=a(rn1)r1

Now, for n = 1

Q(1)=a=a(r(1)1)r1=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+.+arp1=a(rp1)r1 . . . . . . . . (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+.+arp1+ar(p+1)1

Now, using Equation (1):

=a(rp1)r1+arp

=a(rp1)+arp(r1)r1

=a(rp1)+arp