**Prove the following by using the principle of mathematical induction for all n € N:**

**1 + 3 + 3 ^{2}+ ………..+3^{n-1} = (3^{n }– 1)/2**

Solution:

Let the given statement be P(n) , i.e.,

P(n): 1 + 3 + 3^{2}+ ………..+3^{n-1} = (3^{n }– 1)/2

For n = 1, we have

P(1): (3^{1 }– 1)/2 = ^{3-1}/_{2} = ^{2}/_{2} = 1 , which is true.

Let P(k) be true for some positive integer k, i.e.,

1 + 3 + 3^{2}+ ………..+3^{k-1} = (3^{k }– 1)/2

We shall prove that P(k+1) is true.

Consider

1 + 3 + 3^{2}+ ………..+3^{n-1} + 3^{(k+1)-1} = 1 + 3 + 3^{2}+ ………..+3^{k-1} + 3^{k }

=(3^{k }– 1)/_{2} +3^{k}

=[(3^{k }– 1)+2.3^{k}]/_{2}

=[(1+2)3^{k} – 1]/_{2}

= [3.3^{k} – 1]/_{2}

= [3^{k+1} – 1]/_{2}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

**Prove the following by using the principal of mathematical induction for all n €N**

**1 ^{3} + 2^{3} + 3^{3 }+ … + n^{3} = [n(n+1)/_{2}]^{2}**

Solution:

Let the given statement be P(n), i.e.,

P(n): 1^{3} + 2^{3} + 3^{3 }+ … + n^{3} = [n(n+1)/_{2}]^{2}

For n = 1, we have

P(1): 1^{3} = 1 = (^{1(1+1)}/_{2})^{2} = (^{1.2}/_{2})^{2} = 1^{2} = 1, which is true.

Let P(k) be true for some positive integer k, i.e.,

1^{3} + 2^{3} + 3^{3 }+ … + k^{3} = [k(k+1)/_{2}]^{2} —————–(1)

Let us prove that P(k+1) is true.

Consider

1^{3} + 2^{3} + 3^{3 }+ … + k^{3} +(k+1)^{3} = (1^{3} + 2^{3} + 3^{3 }+ … + k^{3})+(k+1)^{3}

Thus, P(k+1) is true whenever P(k) is true.

Hence, by the principal of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

**Prove the following by using the principal of mathematical induction for all n€N**

**1 + ^{1}/_{(1+2)} + ^{1}/_{(1+2+3)} + … + ^{1}/_{(1+2+3+ … +n)} = ^{2n}/_{(n+1)}**

Solution:

Let the given statement be P(n) i.e.,

P(n): 1 + ^{1}/_{(1+2)} + ^{1}/_{(1+2+3)} + … + ^{1}/_{(1+2+3+ … +n)} = ^{2n}/_{(n+1)}

For n = 1, we have

P(1): 1 = ^{2(1)}/_{(1+1)} = ^{2}/_{2} = 1. which is true.

Let P(k) be true for some positive integer k, i.e.,

1 + ^{1}/_{(1+2)} + ^{1}/_{(1+2+3)} + … + ^{1}/_{(1+2+3+ … +k)} = ^{2k}/_{(k+1)} —————-(1)

We shall now prove that P(k+1) is true.

Consider

1 + ^{1}/_{(1+2)} + ^{1}/_{(1+2+3)} + … + ^{1}/_{(1+2+3+ … +k)} + ^{1}/_{(1+2+3+…+k+(k+1)) }= (1 + ^{1}/_{(1+2)} + ^{1}/_{(1+2+3)} + … + ^{1}/_{(1+2+3+ … +k)})+ ^{1}/_{(1+2+3+…+k+(k+1))}

= ^{2k}/_{k+1 }+ ^{1}/_{(1+2+3+…+k+(k+1))} [using(i)]

Thus,P(k+1) is true whenever P(k) is true.

Hence by the principal of mathematical induction, statement P(n) is true for all natural numbers ie., N

**Prove the following by using the principal of mathematical induction for all n€N**

**1.2.3 + 2.3.4 + … + n(n+1)(n+2) = ^{n(n+1)(n+2)(n+3)}/_{4}**

Solution:

Let the given statement be P(n) i.e.,

P(n): 1.2.3 + 2.3.4 + … +n(n+1)(n+2) = ^{n(n+1)(n+2)(n+3)}/_{4}

For n = 1, we have

P(1): 1.2.3 = 6 = ^{1(1+1)(1+2)(1+3)}/_{4} = ^{1.2.3.4}/_{4} = 6, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2.3 + 2.3.4 + … +k(k+1)(k+2) = ^{k(k+1)(k+2)(k+3)}/_{4} ————(i)

We shall now prove that p(k+1) is true.

Consider

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)} + (k + 1) (k + 2) (k + 3)

Thus, P(k+1) is true whenever P(k) is true.

Hence by the principal of mathematical induction, statement P(n) is true for all natural numbers i.e., N

**Prove that the following by using the principal of mathematical induction for all n€N**

**1.3 + 2.3 ^{2} + 3.3^{2} + … + n.3^{n} = {(2n-1)(3^{n+1}+3)} / _{4}**

Solution:

Let the given statement be P(n), i.e.,

**1.3 + 2.3 ^{2} + 3.3^{2} + … + n.3^{n} = {(2n-1)(3^{n+1}+3)} / _{4}**

For n = 1, we have

P(1): 1.3 = 3

Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

**Prove the following by using the principle of mathematical induction for all n N:**

**1.2 + 2.3 + 3.4 + …+n(n+1) = ^{n(n+1)(n+2)}/_{3}**

Solution:

Let the given statement be P(n), i.e.,

P(n) : 1.2 + 2.3 + 3.4 + …+n(n+1) = ^{n(n+1)(n+2)}/_{3}

For n = 1, we have

P(1): 1.2 = 2 = ^{1(1+1)(1+2)}/_{3} = ^{1.2.3}/_{3} = 2 , which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2 + 2.3 + 3.4 + …+k(k+1) = ^{k(k+1)(k+2)}/_{3} ————-(i)

We shall now prove that P(k + 1) is true.

Consider

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

Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

7: Prove the following by using the principle of mathematical induction for all n €N:

1.3 + 3.5 + 5.7 + … + (2n-1)(2n+1) = {n(4n^{2} + 6n – 1)} / _{3}

Solution:

Let the given statement be P(n), i.e.,

P(n): 1.3 + 3.5 + 5.7 + … + (2n-1)(2n+1) = {n(4n^{2} + 6n – 1)} / _{3}

For n = 1, we have

P(1): 1.3 = 3 = {1(4.1 + 6.1 -1)}/3 = (4+6-1)/3 = 9/3 = 3, which is true.

Let P(k) be true for some positive integer k, i.e.,

Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

**Prove the following by using the principle of mathematical induction for all n €N:**

** 1.2 + 2.2 ^{2} + 3.2^{2} + … + n.2^{n} = (n – 1) 2^{n+1} + 2**

Solution:

Let the given statement be P(n), i.e.,

P(n): 1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{n} = (n – 1) 2^{n+1} + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 2^{1+1} + 2 = 0 + 2 = 2, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{k} = (k – 1) 2^{k+1} + 2 ————-(i)

We shall now prove that P(k + 1) is true.

Consider

{1.2 + 2.2^{2} + 3.2^{2} + … + k.2^{k} }+ (k+ 1) 2^{k+1} + 2^{k+1} = (k-1)2^{k+1} + 2+(k+1)2^{k+1}

=2^{k+1}{(k-1)+(k+1)}+2

= 2^{k+1}.2k + 2

=k.2^{(k+1)+1}+2

={(k+1)-1}2^{(k+1)+1}+2

Thus, P(k+1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

9: Prove the following by using the principle of mathematical induction for all n €N:

^{1}/_{2} + ^{1}/_{4} + ^{1}/_{8} + … +^{1}/_{2}^{n} = 1 – ^{1}/_{2}^{n}

Solution:

Let the given statement be P(n), i.e.,

P(n): ^{1}/_{2} + ^{1}/_{4} + ^{1}/_{8} + … +^{1}/_{2}^{n} = 1 – ^{1}/_{2}^{n}

P(1): ^{1}/_{2} = 1 – ^{1}/_{2}^{1} = ^{1}/_{2, }which is true.

**10: Prove the following by using the principle of mathematical induction for all n€ N:**

^{1}/_{2.5} + ^{1}/_{5.8} + ^{1}/_{8.11} + … + ^{1}/_{(3n-1)(3n+2)} = ^{n}/_{(6n+4)}

Solution:

Let the given statement be P(n), i.e.,

P(n):

^{1}/_{2.5} + ^{1}/_{5.8} + ^{1}/_{8.11} + … + ^{1}/_{(3n-1)(3n+2)} = ^{n}/_{(6n+4)}

For n = 1, we have