Mathematical Induction

We have P(n) = n(n + 1).

= P(3) = 3(3 + 1)

= P(3) = 3(4)

Hence, P(3) = 12, So P(3) is also Even.

Given. P(n) = n³ + n is divisible by 3

Find P(3) is true but P(4) is not true

We have P(n) = n³ + n is divisible by 3

Let’s check with P(3)

= P(3) = 3³ + 3

= P(3) = 27 + 3

Therefore P(3) = 30, So it is divisible by 3

Now check with P(4)

= P(4) = 4³ + 4

= P(4) = 64 + 4

Therefore P(4) = 68, So it is not divisible by 3

Hence, P(3) is true and P(4) is not true.

Given. P(n) = “2ⁿ ≥ 3n” and p(r) is true.

Prove. P(r + 1) is true

we have P(n) = 2ⁿ ≥ 3n

Since, P(r) is true So,

= 2^r≥ 3r

Now, Multiply both side by 2

= 2.2^r≥ 3r.2

= 2^{r + 1}≥ 6r

= 2^{r + 1}≥ 3r + 3r [since 3r>3 = 3r + 3r≥3 + 3r]

Therefore 2^{r + 1}≥ 3(r + 1)

Hence, P(r + 1) is true.

Given P(r) is true that means,

= r² + r is even

Let Assume r² + r = 2k - - - - - - (i)

Now, (r + 1)² + (r + 1)

r² + 1 + 2r + r + 1

= (r² + r) + 2r + 2

= 2k + 2r + 2

= 2(k + r + 1)

= 2μ

Therefore, (r + 1)² + (r + 1) is Even.

Hence, P(r + 1) is true

P(n) = 1 + 2 + 3 + - - - - - + n =

P(n) is true for all natural numbers.

Hence, P(n) is true for all n∈N

P(n) = n² - n + 41

= P(1) = 1 - 1 + 41

= P(1) = 41

Therefore, P(1) is Prime

= P(2) = 2² – 2 + 41

= P(2) = 4 - 2 + 41

= P(2) = 43

Therefore, P(2) is prime

= P(3) = 3² – 3 + 41

= P(3) = 9 – 3 + 41

= P(3) = 47

Therefore P(3) is prime

Now, P(41) = (41)² - 41 + 41

= P(41) = 1681

Therefore, P(41) is not prime

Hence, P(1),P(2),P(3) are true but P(41) is not true.

Let us Assume P(n) = 1 + 2 + 3 + - - - - - - + n =

For n = 1

L.H.S of P(n) = 1

R.H.S of P(n) = = 1

Therefore, L.H.S =R.H.S

Since, P(n) is true for n = 1

Let assume P(n) be the true for n = k, so

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

Now

(1 + 2 + 3 + - - + k) + (k + 1)

= + (k + 1)

= (k + 1)

=

=

P(n) is true for n = k + 1

P(n) is true for all n∈N

So , by the principle of Mathematical Induction

Hence, P(n) = 1 + 2 + 3 + - - - + n = is true for all n∈N

Let Assume P(n):1² + 2² + 3² + - - - + n² =

For n = 1

P(1): 1 =

1=1

= P(n) is true for n = 1

Let P(n) is true for n = k, so

P(k): 1² + 2² + 3² + - - - - - + k² =

Let’s check for P(n) = k + 1, So

P(k): 1² + 2² + 3² + - - - - - + k² + (k + 1)² =

= 1² + 2² + 3² + - - - - - + k² + (k + 1)²

=

=

=

=

=

=

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N by PMI

Let P(n) : 1 + 3 + 3² + - - - - + 3^{n - 1} =

Now, For n =1

P(1): 1 = =1

Therefore, P(n) is true for n =1

Now , P(n) is true for n = k

P(k) : 1 + 3 + 3² + - - - - + 3^{k - 1} = - - - - - (1)

Now, We have to show P(n) is true for n = k + 1

i.e P(k + 1): 1 + 3 + 3² + - - - - + 3^k =

then, {1 + 3 + 3² + - - - - + 3^{k - 1}} + 3^{k + 1 - 1}

= using equation (1)

=

=

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N

Let P(n):

For n = 1

P(1):

= P(n) is true for n = 1

Let P(n) is true for n = k, So

- - - - - (1)

Now, Let P(n) is true for n = k + 1, So

Then,

=

=

=

=

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N

Let P(n): 1 + 3 + 5 + … + (2n - 1) = n²

Let check P(n) is true for n = 1

P(1) = 1 =1²

1 = 1

P(n) is true for n = 1

Now, Let’s check P(n) is true for n = k

P(k) = 1 + 3 + 5 + … + (2k - 1) = k² - - - (1)

We have to show that

1 + 3 + 5 + … + (2k - 1) + 2(k + 1) - 1 = (k + 1)²

Now,

= 1 + 3 + 5 + … + (2k - 1) + 2(k + 1) - 1

= k² + (2k + 1)

= k² + 2k + 1

= (k + 1)²

Therefore, P(n) is true for n =k + 1

Hence, P(n) is true for all n∈N.

Let P(n):

Step 1: Let us check if P(1) is true or not,

P(1):

Therefore, P(1) is true.

Step 2: Let us assume that P(k) is true, now we have to prove that P(k + 1) is true.

P(k):

P(k+1):

From P(k) we can see that,

P(k + 1):

P(k + 1):

P(k + 1):

Therefore, P(k + 1) is true.

Hence, Proved by mathematical induction.

Let P(n):

For n= 1 is true,

P(1):

Since, P(n) is true for n = 1

Let P(n) is true for n = k, so

- - - - (1)

We have to show that,

Now,

=

=

=

=

=

=

=

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N

Let P(n):

Step1: Let us verify P(1).

P(1):

P(1):

Therefore, P(1) is true.

Step 2:

Let P(k) is true.

Therefore, P(k):

Now we have to prove that P(k + 1) is also true.

So,

L.H.S =

L.H.S =

Now from P(k) we can say that,

Putting this value, we get,

L.H.S =

L.H.S =

L.H.S =

L.H.S = R.H.S

Hence, Proved.

Let P(n):

For n= 1is true

P(1): =

Since, P(n) is true for n =1

Let P(n) is true for n= k

P(n): - - - - - - - (1)

We have to show that,

Now,

=

=

=

=

=

=

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈ N

Let P(n): 1.2 + 2.2² + 3.2³ + … + n.2ⁿ=(n–1) 2^{n + 1} + 2

For n = 1

= 1.2 = 0.2⁰ + 2

= 2 = 2

Since, P(n) is true for n = 1

Let P(n) is true for n = k, so

P(k): 1.2 + 2.2² + 3.2³ + … + k.2^k=(k–1) 2^{k + 1} + 2 - - - - - - (1)

We have to show that,

{1.2 + 2.2² + 3.2³ + … + k.2^k + (k + 1) 2^{k + 1} = k.2^{k + 2} + 2

Now,

{1.2 + 2.2² + 3.2³ + … + k.2^k} + (k + 1)2^{k + 1}

= [(k - 1)2^{k + 1} + 2] + (k + 1)2^{k + 1} using equation (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 + 2} + 2

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N by PMI

Let P(n): 2 + 5 + 8 + 11 + … + (3n – 1) = n(3n + 1)

For n=1

P(1): 2 = .1.(4)

2 = 2

Since, P(n) is true for n = 1

Let P(n) is true for n = k, so

P(k): 2 + 5 + 8 + 11 + … + (3k – 1) = k(3k + 1) - - - - - - - (1)

We have to show that,

2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2) = (k + 1)(3k + 4)

Now,

{2 + 5 + 8 + 11 + … + (3k – 1)} + (3k + 2)

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

=

=

=

=

=

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n ∈ N by PMI

Let P(n): 1.3 + 2.4 + 3.5 + … + n.(n + 2) =

For n = 1

P(1): 1.3 = .1.(2)(9)

= 3 = 3

Since, P(n) is true for n = 1

Now,

For n = k

= P(n): 1.3 + 2.4 + 3.5 + … + k . (k + 2)= - - - - - (1)

We have to show that

= 1.3 + 2.4 + 3.5 + … + k . (k + 2) + (k + 3) =

Now,

= {1.3 + 2.4 + 3.5 + … + k (k + 2)} + (k + 1)(k + 3)

= k(k + 1)(2k + 7) + (k + 1)(k + 3) using equation (1)

= (k + 1)

= (k + 1)

= (k + 1)

= (k + 1)

= (k + 1)

= (k + 1)

= (k + 1)(k + 2)(2k + 9)

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈ N

Let P(n): 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1)

For n = 1

P(1): (2.1 – 1) (2.1 + 1) =

= 1x3 =

= 3 = 3

Since, P(n) is true for n =1

Now, For n = k, So

1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) - - - - - - - (1)

We have to show that,

1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1)(2k + 3)

Now,

1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1)(2k + 3)

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

=

=

=

=

=

=

Therefore, P(n) is true for n=k + 1

Hence, P(n) is true for all n∈ N by PMI

Let P(n): 1.2 + 2.3 + 3.4 + … + n(n + 1)=

For n = 1

P(1): 1(1 + 1)=

= 1x2 =

= 2 = 2

Since, P(n) is true for n = 1

Let P(n) is true for n = k

= P(k): 1.2 + 2.3 + 3.4 + … + k(k + 1)= - - - - - (1)

We have to show that,

= 1.2 + 2.3 + 3.4 + … + k(k + 1) + (k + 1)(k + 2)=

Now,

{1.2 + 2.3 + 3.4 + … + k(k + 1)} + (k + 1)(k + 2)

=

= (k + 2)(k + 1)

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n ∈ N

Let P(n):

For n =1 is true,

P(1):

=

Since, P(n) is true for n =1

Now, For n = k

P(k): - - - - (1)

We have to show that,

Now,

= using equation (1)

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n ϵ N by PMI

Let P(n): 1² + 3² + 5² + … + (2n – 1)² =

For n = 1

= (2.1 – 1)² =

= 1 = 1

Since, P(n) is true for n = 1

Let P(n) is true for n = k ,

P(k) ): 1² + 3² + 5² + … + (2k – 1)² = - - (1)

We have to show that,

1² + 3² + 5² + … + (2k – 1)² + (2k + 1)² =

Now,

{1² + 3² + 5² + … + (2k – 1)²} + (2k + 1)²

= using equation (1)

=

= (2k + 1)

= (2k + 1)

= (2k + 1)

=

=

=

=

=

=

Therefore, P(n)is true for n = k + 1

Hence, P(n) is true for all n∈ N

Let P(n): a + ar + ar² + … + ar^{n - 1} =

For n =1

a = a

a = a

Since, P(n) is true for n = 1

Let P(n) is true for n = k , so

P(k): a + ar + ar² + … + ar^{k - 1} = - - - - - - - (1)

We have to show that,

a + ar + ar² + … + ar^{k - 1} + ar^k =

Now,

{ a + ar + ar² + … + ar^{k - 1}} + ar^k

= using equation (1)

=

=

=

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n ∈ N

P(n): a + (a + d) + (a + 2d) + … + (a + (n– 1)d) =

For n = 1

a = [2a + (1 - 1)d]

a = a

Since, P(n) is true for n =1,

Let P(n) is true for n = k, so

a + (a + d) + (a + 2d) + … + (a + (k– 1)d) = - - - - - (1)

We have to show that,

a + (a + d) + (a + 2d) + … + (a + (k– 1)d) + (a + (k)d) = [2a + kd]

Now,

{a + (a + d) + (a + 2d) + … + (a + (k– 1)d)} + (a + kd)

= [2a + (k - 1)d] + (a + kd) using equation

=

=

=

=

= [2a + kd]

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true all n∈ N by PMI

Let P(n): 5²ⁿ - 1 is divisible by 24

Let’s check For n =1

P(1): 5² - 1 = 25 - 1

= 24

Since, it is divisible by 24

So, P(n) is true for n=1

Now, for n=k

5^2k - 1 is divisible by 24

P(k): 5^2k - 1 = 24λ - - - - - - - (1)

We have to show that,

5^{2k + 1} - 1 is divisible by 24

5^{2(k + 1)} - 1 = 24μ

Now,

5^{2(k + 1)} - 1

= 5^2k.5^{2 -} 1

= 25.5^2k - 1

= 25.(24λ + 1) - 1 using equation (1)

= 25.24λ + 24

= 24λ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N by PMI

Let P(n): 3²ⁿ + 7 is divisible by 8

Let’s check For n =1

P(1): 3² + 7 = 9 + 7

= 16

Since, it is divisible by 8

So, P(n) is true for n=1

Now, for n=k

P(k): 3^2k + 7 = 8λ - - - - - - - (1)

We have to show that,

3^{2(k + 1)} + 7 is divisible by 8

3^{2k + 2} + 7 = 8μ

Now,

3^{2(k + 1)} + 7

= 3^2k.3² + 7

= 9.3^2k + 7

= 9.(8λ - 7) + 7

= 72λ - 63 + 7

= 72λ - 56

= 8(9λ - 7)

= 8μ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N by PMI

Let P(n): 5^{2n + 2} – 24n – 25

For n = 1

= 5^{2.1 + 2} - 24.1 - 25

= 625 – 49

= 576

Since, it is divisible by 576

Let P(n) is true for n=k, so

= 5^{2k + 2} – 24k – 25 is divisible by 576

= 5^{2k + 2} – 24k – 25 = 576λ - - - - - (1)

We have to show that,

= 5^{2k + 4} – 24(k + 1) – 25 is divisible by 576

= 5^{(2k + 2) + 2} – 24(k + 1) – 25 = 576μ

Now,

= 5^{(2k + 2) + 2} – 24(k + 1) – 25

= 5^{(2k + 2)}.5² – 24k – 24– 25

= (576λ + 24k + 25)25 – 24k– 49 using equation (1)

= 25. 576λ + 576k + 576

= 576(25λ + k + 1)

= 576μ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N by PMI

Let P(n): 3^{2n + 2} – 8n – 9

For n = 1

= 3^{2.1 + 2} - 8.1 - 9

= 81 – 17

= 64

Since, it is divisible by 8

Let P(n) is true for n=k, so

= 3^{2k + 2} – 8k – 9 is divisible by 8

= 3^{2k + 2} – 8k – 9 = 8λ - - - - - (1)

We have to show that,

= 3^{2k + 4} – 8(k + 1) – 9 is divisible by 8

= 3^{(2k + 2) + 2} – 8(k + 1) – 9 = 8μ

Now,

= 3^{2(k + 1)}.3² – 8(k + 1) – 9

= (8λ + 8k + 9)9 – 8k – 8 – 9

= 72λ + 72k + 81 - 8k - 17 using equation (1)

= 72λ + 64k + 64

= 8(9λ + 8k + 8)

= 8μ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N by PMI

Let P(n) : (ab)ⁿ = aⁿ bⁿ

Let check for n = 1 is true

= (ab)¹ = a¹b¹

= ab = ab

Therefore, P(n) is true for n =1

Let P(n) is true for n=k,

= (ab)^k=a^k.b^k - - - - - - (1)

We have to show that,

= (ab)^{k + 1}=a^{k + 1}.b^{k + 1}

Now,

= (ab)^{k + 1}

=(ab)^k (ab)

= (a^kb^k)(ab) using equation (1)

= (a^{k + 1})(b^{K + 1})

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N by PMI

Let P(n): n(n + 1) (n + 5) is multiple by 3 for all n∈N

Let P(n) is true for n=1

P(1): 1(1 + 1) (1 + 5)

= 2 × 6

= 12

Since, it is multiple of 3

So, P(n) is true for n = 1

Now, Let P(n) is true for n = k

P(k): k(k + 1) (k + 5)

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

Then, k(k + 1) (k + 5) = 3λ - - - - - (1)

We have to show,

= (k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

= (k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ

Now,

= (k + 1)[(k + 1) + 1][(k + 1) + 5]

= (k + 1)(k + 2)[(k + 1) + 5]

= [k(k + 1) + 2(k + 1)][(k + 5) + 1]

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

= 3λ + k² + k + 2(k² + 6k + 5) + 2k + 2

= 3λ + k² + k + 2k² + 12k + 10 + 2k + 2

= 3λ + 3k² + 15k + 12

= 3(λ + k² + 5k + 4)

= 3μ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n∈N

Let P(n): 7²ⁿ + 2^{3n - 3}.3^{n - 1} is divisible by 25

For n=1

= 7² + 2⁰.3⁰

= 49 + 1

= 50

Therefor it is divisible by 25

So, P(n) is true for n = 1

Now, P(n) is true For n = k,

So, we have to show that 7²ⁿ + 2^{3n - 3}.3^{n - 1} is divisible by 25

= 7^2k + 2^{3k - 3}.3^{k - 1} = 25λ - - - - - - - (1)

Now, P(n) is true For n = k + 1,

So, we have to show that 7^{2k + 1} + 2^3k.3^k is divisible by 25

= 7^{2k + 2} + 2^3k.3^k = 25μ

Now,

= 7^{2(k + 1)} + 2^3k.3^k

= 7^2k.7¹ + 2^3k.3^k

= (25λ – 2^{3k - 3}.3^{k - 1})49 + 2^3k.3k from eq 1

=

= 24×25×49λ - 2^3k.3^k..49 + 24.2^3k.3^k

= 24×25×49λ - 25.2^3k.3^k

= 25(24.49λ - 2^3k.3^k)

= 25μ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n ∈ N

Let P(n) = 2.7ⁿ + 3.5ⁿ – 5

Now, P(n): 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 for all n ϵ N

Step1:

P(1) = 2.7 + 3.5 – 5 = 1.2

Thus, P(1) is divisible by 24

Step2:

Let, P(m) be divisible by 24

Then, 2.7^m + 3.5^m – 5 = 24λ, where λ ϵ N.

Now, we need to show that P(m+1) is true whenever P(m) is true.

So, P(m+1) = 2.7^m+1 + 3.5^m+1 – 5

= 2.7^m+1 + 5.( 2.7^m + 3.5^m – 5 ) – 5

= 2.7^m+1 + 5.( 24λ + 5 - 2.7^m ) – 5

= 2.7^m+1 + 120λ + 25 - 10.7^m – 5

= 2.7^m.7 - 10.7^m+ 120 λ +24 – 4

= 7^m(14 – 10) + 120 λ +24 – 4

= 7^m(4) + 120 λ +24 – 4

= 7^m(4) + 120 λ +24 – 4

= 4(7^m - 1) + 24(5λ +1)

As, 7^m – 1 is a multiple of 6 for all m ϵ N.

So, P(m+1) = 4.6μ +24(5λ +1)

= 24(μ +5λ +1)

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all n ϵN.

Let P(n) = 11ⁿ⁺² + 12²ⁿ⁺¹

Now, P(n): 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133 for all n ϵ N

Step1:

P(1) = 1331 + 1728 = 3059

Thus, P(1) is divisible by 133

Step2:

Let, P(m) be divisible by 24

Then, 11^m+2 + 12^2m+1 = 133λ, where λ ϵ N.

Now, we need to show that P(m+1) is true whenever P(m) is true.

So, P(m+1) = 11^m+3 + 12^2m+3

= 11^m+2.11+ 12^2m+1.12²+11.12^2m+1 – 11.12^2m+1

= 11.(11^m+2+ 12^2m+1) + 12^2m+1(144-11)

= 11.133λ + 12^2m+1.133

= 133.(11 λ + 12^2m+1)

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all n ϵN.

Let P(n) = 1×1! + 2×2! + 3×3! +…+ n×n

P(n): 1×1! + 2×2! + 3×3! +…+ n×n! = (n + 1)! – 1 for all n ϵ N

Step1:

P(1) = 1×1! = (2)! – 1 = 1

Thus, P(n) is equal to (n + 1)! – 1 for n = 1

Step2:

Let, P(m) be equal to (m + 1)! – 1

Then, 1×1! + 2×2! + 3×3! +…+ m×m! = (m + 1)! – 1

Now, we need to show that P(m+1) is true whenever P(m) is true.

P(m+1) = 1×1! + 2×2! + 3×3! +…+ m×m! + (m+1)×(m+1)!

= (m+1)! – 1 + (m+1)×(m+1)!

= (m+1)!(m+1+1) – 1

= (m+1)!(m+2) – 1

= (m+2)! – 1

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all nϵN.

Let P(n) = n³ – 7n + 3

Now, P(n): n³ – 7n + 3 is divisible by 3 for all n ϵ N

Step1:

P(1) = 1 – 7 + 3 = -3

Thus, P(1) is divisible by 3

Step2:

Let, P(m) be divisible by 24

Then, n³ – 7n + 3 = 3λ, where λ ϵ N.

Now, we need to show that P(m+1) is true whenever P(m) is true.

So, P(m+1) = (n+1)³ – 7(n+1) + 3

= n³+3n²+3n+1-7n-7+3

= n³– 7n + 3 +3n²+3n+1-7

= 3λ+3(n²+n-2)

= 3(λ+n²+n-2)

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all n ϵN.

Let P(n) = 1 + 2 + 2² + … +

P(n): 1 + 2 + 2² + … + 2ⁿ = 2^{n + 1} – 1 for all n ϵ N

Step1:

P(1) = 1 = (2) – 1 = 1

Thus, P(n) is equal to 2^{n + 1} – 1 for n = 1

Step2:

Let, P(m) be equal to 2^{m + 1} – 1

Then, 1 + 2 + 2² + … + 2^m = 2^{m + 1} – 1

Now, we need to show that P(m+1) is true whenever P(m) is true.

P(m+1) = 1 + 2 + 2² + … + 2^m + 2^{m + 1}

= 2^{m + 1} – 1 + 2^{m + 1}

= 2.2^{m + 1} –1

= 2^{m + 2} – 1

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all nϵN.

Let P(n) = 7 + 77 + 777 + … + 777……n times……7

Step1:

Step2:

Now, we need to show that P(m+1) is true whenever P(m) is true.

This is a geometric progression with n = m+1

So, P(m+1) = 7 + 77 + 777 + … + 777……(m+1) times……7

Thus, P(m+1) is true.

So, by principle of mathematical induction, P(n) is true for all nϵN.

Step1:

Step2:

Now, we need to show that P(m+1) is true whenever P(m) is true.

It is a positive integer.

Thus, P(m+1) is true.

So, by principle of mathematical induction, P(n) is true for all nϵN.

Step1:

Step2:

Now, we need to show that P(m+1) is true whenever P(m) is true.

It is a positive integer.

Thus, P(m+1) is true.

So, by principle of mathematical induction, P(n) is true for all nϵN.

Step1: For n=1

So, it is true for n=1

Step2:

Now, we need to show that P(m+1) is true whenever P(m) is true.

Now,

Thus, P(m+1) is true.

Let P(n) =

Let us find if it is true at n = 2,

P(2):

P(2):

Hence, P(2) holds.

Now let P(k) is true, and we have to prove that P(k + 1) is true.

Therefore, we need to prove that,

P(k) = …….(1)

Taking L.H.S of P(k) we get,

P(k) =

P(k + 1) =

From equation (1),

P(k + 1) =

P(k + 1) =

P(k + 1) =

P(k + 1) =

Therefore, P(k + 1) holds.

Hence, P(n) is true for all n ≥ 2.

Step1:

Thus, P(1)is true.

Step2:

Now, we need to show that P(m+1) is true whenever P(m) is true.

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all nϵN.

Let the given statement be P(n)

Thus, P(2) is true.

Let, P(m) be true,

Now,

Now, we need to prove that P(m+1) is true whenever P(m) is true.

Thus, Pm+1 is true. By the principle of mathematical induction, P(n) is true for all n∈N, n≥2.

Let, P(n) be the given statement,

Now, P(n):x^2n-1 + y^{2n – 1}

Step1: P(1):x+y which is divisible by x+y

Thus, P(1) is true.

Step2: Let, P(m) be true.

Then, x^2m-1+y^2m-1= λ(x+y)

Now, P(m+1) = x^2m+1+y^2m+1

= x^2m+1+y^2m+1-x^2m-1.y²+x^2m-1.y²

= x^2m-1(x²-y²) + y²(x^2m-1+y^2m-1)

= (x+y)(x^2m-1(x-y)+λy²)

Thus, P(m+1) is divisible by x+y. So, by the principle of mathematical

induction P(n) is true for all n.

Let, P(n) be the given statement,

Thus, P(1) is true.

Step2: Let, P(m) be true.

Now, we need to show that P(m+1) is true when P(m) is true.

As P(m) is true

Thus, P(m+1) is divisible by x+y. So, by the principle of mathematical

induction P(n) is true for all n.

Step1: For n=1

L.H.S = cos [α+(1-1)β] = cos α

As, L.H.S = R.H.S

So, it is true for n=1

Step2: For n=k

Now, we need to show that P(k+1) is true when P(k) is true.

Adding cos(α+kβ) both sides of P(k)

As, LHS = RHS

Thus, P(k+1) is true. So, by the principle of mathematical induction

P(n) is true for all n.

Step1: For n=2

So, it is true for n=2

Step2: For n=k

Now, we need to show that P(k+1) is true when P(k) is true.

As, LHS = RHS

Thus, P(k+1) is true. So, by the principle of mathematical induction

P(n) is true for all n.

Step1: For n=1

As LHS=RHS.

So, it is true for P(1)

For n=k, let P(k) be true.

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

As L.H.S=R.H.S

Thus, P(k+1) is true. So, by the principle of mathematical induction

P(n) is true for all n.

If P(r) is true then 2^r ≥ 3r

For, P(r+1)

2^r+1=2.2^r

For, x>3, 2x>x+3

So, 2.2^r>2^r+3 for r>1

⇒ 2^r+1>2^r+3 for r>1

⇒ 2^r+1>3r +3 for r>1

⇒ 2^r+1>3(r+1) for r>1

So, if P(r) is true, then P(r+1) is also true.

For, n=1, P(1):

L.H.S=2

R.H.S=3

As L.H.S<R.H.S

So, it is not true for n=1

Hence, P(n) is not true for all natural numbers.

Step1: For n=1, P(1):

LHS=S₁=1

RHS=S₁=1

So, P(1) is true.

Step2: Let P(n) be true for n=k

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

Case1: When k is odd, then (k+1) is even

As LHS=RHS

So, it is true for n=k+1 when k is odd.

Case2: When k is even, then (k+1) is odd

As LHS=RHS

So, it is true for n=k+1 when k is even.

Hence, by the principle of mathematical induction P(n) is true ∀ nϵN.

Let the given statement be defined as

P(n): The number of subsets of a set containing n distinct

elements=2ⁿ, for all n ϵ N.

Step1: For n=1,

L.H.S=As, the subsets of the set containing only 1 element are:

Φ and the set itself

i.e. the number of subsets of a set containing only element=2

R.H.S=2¹=2

As, LHS=RHS, so, it is true for n=1.

Step2: Let the given statement be true for n=k.

P(k): The number of subsets of a set containing k distinct

elements=2^k

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

Let A={a₁, a₂, a₃, a₄,…, a_k, b} so that A has (k+1) elements.

So the subset t of A can be divided into two collections:

first contains subsets of A which don t have b in them and

the second contains subsets of A which do have b in them.

First collection: { }, {a₁},{a₁, a₂},{a₁, a₂, a₃},…,{a₁, a₂, a₃, a₄,…, a_k} and

Second collection: {b}, {a₁,b},{a₁,a₂,b },{a₁,a₂,a₃,b},…,{a₁,a₂,a₃,a₄,…,a_k, b}

It can be clearly seen that:

The number of subsets of A in first collection

= The number of subsets of set with k elements i.e. { a₁, a₂, a₃, a₄,…, a_k}=2^k

Also it follows that the second collection must have

the same number of the subsets as that of the first = 2^k

So the total number of subsets of A=2^k+2^k=2^k+1

Thus, by the principle of mathematical induction P(n) is true.

Let P(n): a_n=3.7^n-1 for all n ϵ N

Step1: For n=1,

a₁=3.7^1-1=3

So, it is true for n=1

Step2: For n=k,

Let P(k) be true.

So, a_k=3.7^k-1

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

a_k+1=7.a_k

=7.3.7^k-1

=3.7^k-1+1

=3.7^(k+1)-1

So, it is true for n=k+1

Hence, by the principle of mathematical induction P(n) is true.

Step1: For n=1

So, it is true for n=1.

Step2: For n=k,

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

So, it is true for n=k+1.

Thus, by the principle of mathematical induction P(n) is true.

Let P(n): x_n =5+4n for all nϵN

Step1: For n=0,

P(0):x₀=5+4×0=5

So, it is true for n=0.

Step2: Let P(k) be true

Thus, x_k =5+4k

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

x_k+1 =4+ x_k+1-1

=4+x_k

=4+5+4k

=5+4(k+1)

=RHS

Thus, P(k+1) is true, so by mathematical induction P(n) is true.

Step1: For n=2, P(n):

Therefore, it is true for n=2.

Step2: Let P(n) be true for n=k.

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

So, it is true for n=k+1, thus by the principle of mathematical induction P(n) is true for all n ≥ 2

Let P(n):c(a₁+a₂+…+a_n) = ca₁+ca₂+…+ca_n ,for all natural

numbers, n ≥ 2.

Step1: For n=2,

P(2)

LHS= c(a₁ + a₂)

RHS= c a₁ + ca₂

As, it is given that c(a₁ + a₂) = c a₁ + ca₂

Thus, P(2) is true.

Step2: For n=k,

Let P(k) be true

So, c(a₁+a₂+…+a_k) = ca₁+ca₂+…+ca_k

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

LHS= c(a₁+a₂+…+a_K+ak₊₁)

=c[(a₁+a₂+…+a_K)+a_k+1]

=c(a₁+a₂+…+a_K)+ca_k+1

=ca₁+ca₂+…+ca_K+ca_k+1

=RHS

Thus, P(k+1) is true, so by mathematical induction P(n) is true.

The first principle of mathematical induction states that if the basis step and the inductive step are proven, then P(n) is true for all natural numbers.

The set of value of n for which the statement P(n): 2n < n! is true can be written as {n ∈ N : n ≥ 4}.

Let M be an integer. Suppose we want to prove that P(n) is true for all positive integers ≥M. Then if we show that:

Step 1: P(M) is true, and

Step 2: for an arbitrary positive integer k≥M, if P(M).P(M+1).P(M+2)……P(k) are true then P(k+1) is true,

Then P(n) is true for all positive integers greater than or equal to M.

for n=1,

2×4^2×1+1 + 3^3×1+1=2×4³+3⁴

= 2×64+81

= 128+81

= 209

For n=2,

2×4^2×2+1 + 3^3×2+1 = 2×4⁵+3⁷

= 2×1024+2187

= 2048+2187

= 4235

Now, the H.C.F of 209 and 4235 is 11.

Hence, λ=11.

Given xⁿ-1 is divisible by x-λ

⇒ x=λ is the root of the eqn xⁿ-1

⇒λⁿ-1=0

⇒λⁿ=1

Least value of λ=1

Given for all n€ N 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹

For n=1,

3 × 5³ + 2⁴

3 × 125 + 16

375 + 16 = 391

For n=2,

3 × 5⁵ + 2⁷

3 × 3125 +128

9375 + 128 = 9503

H.C.F of 391, 9503 = 17

Given 10ⁿ+3 ×4ⁿ⁺²+λ is divisible by 9

For n=1,

10 + 3 × 4³ + λ

10 + 3 × 64 + λ

= 202 + λ

202 when divided by 9 gives remainder 4

For n=2,

10² + 3 × 4⁴ + λ

=100 + 3 × 256 + λ

=868 + λ

868 when divided by 9 gives remainder 4

؞ λ = 4 + 1 = 5

Given P(n):2n< (1×2×…. ×n)

For n=1, 2<2

For n=2, 4<4

For n=3, 6<6

For n=4, 8<24

the smallest positive integer for which P(n) is true is 4.

Since given P(5) is true and P(k) is true for all k>5€N,

then we can conclude that P(n) is true for all n≥5

For n=1,

49 16 + λ

⇒ 65 + λ

Now we can see that if λ = -1, then it is divisible by 64

؞ λ = -1