Principle Of Mathematical Induction

To Prove:

1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1)

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n): 1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1)

Step 1:

P(1) = 1/2 1(1 + 1) = 1/2 × 2 = 1

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

P(k): 1 + 2 + 3 + 4 + … + k = 1/2 k(k + 1)

Now,

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

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

= 1/2 (k + 1) (k + 2)

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1) for all n ϵ N

Hence proved.

To Prove:

2 + 4 + 6 + 8 + …. + 2n = n(n + 1)

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n): 2 + 4 + 6 + 8 + …. + 2n = n(n + 1)

Step 1:

P(1) = 1(1 + 1) = 1 × 2 = 2

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

P(k): 2 + 4 + 6 + 8 + …. + 2k = k(k + 1)

Now,

2 + 4 + 6 + 8 + …. + 2k + 2(k + 1) = k(k + 1) + 2(k + 1)

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

= (k + 1) (k + 2)

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

2 + 4 + 6 + 8 + … + 2n = n(n + 1) for all n ϵ N

To Prove:

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n):

Step 1:

P(1) = = 1

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

P(k):

Now,

=

=

=

=

=

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

for all n ϵ N

To Prove:

2 + 6 + 18 + … + 23^n–1 = (3ⁿ –1)

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n): 2 + 6 + 18 + … + 2 × 3^n–1 = (3ⁿ –1)

Step 1:

P(1) = 3¹ –1 = 3 - 1 = 2

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

P(k): 2 + 6 + 18 + … + 23^k–1 = (3^k –1)

Now,

2 + 6 + 18 + … + 2 × 3^k–1 + 2 × 3^{k + 1–1} = (3^k –1) + 2 × 3^k

= - 1 + 3 × 3^k

= 3^{k + 1 -} 1

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

2 + 6 + 18 + … + 23^n–1 = (3ⁿ –1) for all n ϵ N

To Prove:

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n):

Step 1:

P(1) =

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

P(k):

Now,

=

=

=

=

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

for all n ϵ N

To Prove:

1² + 3² + 5² + 7² + … + (2n – 1)² =

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n): 1² + 3² + 5² + 7² + … + (2n – 1)² =

Step 1:

P(1) = = 1

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

P(k): 1² + 3² + 5² + 7² + … + (2k – 1)² =

Now,

1² + 3² + 5² + 7² + … + (2(k + 1)–1)² =

=

=

=

=

= (Splitting the middle term)

⁼

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

1² + 3² + 5² + 7² + … + (2n – 1)² = for all n ϵ N

To Prove:

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):

For n = 1

LHS = 1 × 2 = 2

RHS = (1 - 1) × 2^{(1 + 1)} + 2

= 0 + 2 = 2

Hence, LHS = RHS

P(n) is true for n 1

Assume P(k) is true

……(1)

We will prove that P(k + 1) is true

1×

1×

1× ……(2)

We have to prove P(k + 1) from P(k), i.e. (2) from (1)

From (1)

1×

Adding both sides,

(1× +

=

=

=

(1× =

which is the same as P(k + 1)

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for

where n is a natural number

Put k = n - 1

(1× =

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):

For n = 1

LHS = = 12

RHS =

= = 12

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

……(1)

We will prove that P(k + 1) is true

…(2)

We have to prove P(k + 1) from P(k) ie (2) from (1)

From (1)

Adding both sides

which is the same as P(k + 1)

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

By the principle of mathematical induction, P(n) is true for×

where n is a natural number

Put k = n - 1

Hence proved

To Prove:

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):

For n = 1

LHS =

RHS = 1

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

……(1)

We will prove that P(k + 1) is true

RHS =

LHS =

[ Writing the last

Second term ]

= [From 1]

{ 1 + 2 + 3 + 4 + … + n = [n(n + 1)]/2 put n = k + 1 }

=

[ Taking LCM and simplifying ]

=

= RHS

Therefore ,

LHS = RHS

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

By the principle of mathematical induction, P(n) is true for×

where n is a natural number

Put k = n - 1

Hence proved

To Prove:

For n = 1

LHS =

RHS =

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

……(1)

We will prove that P(k + 1) is true

RHS =

LHS =

[ Writing the Last second term ]

=

= [ Using 1 ]

=

=

= ( Taking LCM and simplifying )

=

= RHS

Therefore, =

LHS = RHS

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for

where n is a natural number

Put k = n - 1

=

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):

For n = 1

LHS =

RHS =

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

= ……(1)

We will prove that P(k + 1) is true

RHS =

LHS =

=

[ Writing the second last term ]

= [ Using 1 ]

=

=

=

( Splitting the numerator and cancelling the common factor)

= RHS

LHS = RHS

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

By the principle of mathematical induction, P(n) is true for

where n is a natural number

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):

For n = 1

LHS =

RHS =

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

= ……(1)

We will prove that P(k + 1) is true

RHS =

LHS =

=

[ Writing the second last term ]

= [ Using 1 ]

=

=

=

( Splitting the numerator and cancelling the common factor)

= RHS

LHS = RHS

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for×

where n is a natural number

Hence proved.

To Prove:

For n = 1

LHS =

RHS =

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

……(1)

We will prove that P(k + 1) is true

RHS =

LHS = [ Writing the Last second term ]

=

= [ Using 1 ]

=

=

= ( Taking LCM and simplifying )

=

= RHS

Therefore, =

LHS = RHS

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

By the principle of mathematical induction, P(n) is true for

where n is a natural number

Put k = n - 1

=

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):

For n = 1

LHS =

RHS = = 4

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

= …(1)

We will prove that P(k + 1) is true

RHS =

LHS =

[Now writing the second last term ]

=

= [ Using 1 ]

=

=

=

=

=

= RHS

LHS = RHS

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for

where n is a natural number

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):

For n = 1

LHS =

RHS = = 2

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

= …(1)

We will prove that P(k + 1) is true

RHS =

LHS =

[Now writing the second last term ]

=

= [ Using 1 ]

=

=

= k + 2

= RHS

LHS = RHS

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

By the principle of mathematical induction, P(n) is true for

where n is a natural number

Hence proved.

To Prove:

n × ( n + 1 ) × ( n + 2 ) is multiple of 6

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

n × ( n + 1 ) × ( n + 2 ) is multiple of 6

Let P(n): , which is multiple of 6

For n = 1 P(n) is true since , which is multiple of 6

Assume P(k) is true for some positive integer k , ie,

= k × ( k + 1 ) × ( k + 2 ) = 6m , where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider ,

= (k + 1) × (( k + 1) + 1 )× ( (k + 1) + 2 )

= (k + 1)×{ k + 2 }×{ (k + 2) + 1 }

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

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

= [6m + 2×(k + 1)×(k + 2) ] + (k + 1)×(k + 2)

= 6m + 3×(k + 1)×(k + 2)

Now , (k + 1) & (k + 2) are consecutive integers , so their product is even

Then , (k + 1)×(k + 2) = 2×w (even)

Therefore ,

= 6m + 3×[ 2×w ]

= 6m + 6×w

= 6(m + w)

= 6×q where q = ( m + w ) is some natural number

Therefore

is multiple of 6

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

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n):

For n = 1 P(n) is true since

which is divisible by x + y

Assume P(k) is true for some positive integer k , ie,

=

, where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider ,

[ Adding and subtracting ]

[ Using 1 ]

, which is factor of ( x + y )

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers ie, N

Hence proved

To Prove:

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n):

For n = 1

P(n) is true since

which is divisible by x - 1

Assume P(k) is true for some positive integer k , ie,

=

, where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider ,

[ Adding and subtracting 1 ]

[ Using 1 ]

, which is factor of ( x - 1 )

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n):

For n = 1 P(n) is true since

which is multiple of 27

Assume P(k) is true for some positive integer k , ie,

=

, where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider ,

= [ Adding and subtracting ]

[ Using 1 ]

, where r = is a natural number

Therefore is divisible of 27

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n):

For n = 1 P(n) is true since

which is divisible of 9

Assume P(k) is true for some positive integer k , ie,

=

, where m ∈ N …(1)

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

Consider ,

[ Adding and subtracting

]

[ Using 1 ]

, where r = is a natural number

Therefore

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.

To Prove:

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n):

For n = 1 P(n) is true since

which is divisible of 8

Assume P(k) is true for some positive integer k , ie,

=

, where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider ,

[ Adding and subtracting 8k + 9 ]

= 9(8m) + 72k + 81 -8k-17 [ Using 1 ]

= 9(8m) + 64k + 64

= 8(9m + 8k + 8)

= 8×r , where r = 9m + 8k + 8 is a natural number

Therefore is a divisible of 8

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.

To Prove:

, which is multiple of 7

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

is multiple of 7

Let P(n): , which is multiple of 7

For n = 1 P(n) is true since , which is multiple of 7

Assume P(k) is true for some positive integer k , ie,

= , where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider ,

[Adding and subtracting ]

[ Using 1 ]

, where r = is a natural number

Therefore is multiple of 7

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers ie, N

Hence proved

To Prove:

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n):

For n = 1 P(n) is true since , which is true

Assume P(k) is true for some positive integer k , ie,

= …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider ,

[ Using 1 ]

[Multiplying and dividing by 2 on RHS ]

Now ,

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.