Principles of Mathematical Induction MCQ Quiz - Objective Question with Answer for Principles of Mathematical Induction - Download Free PDF

Concept:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P(1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P (k) implies the truth of P (k + 1).

Then, P (n) is true for all natural numbers n

Calculation:

We have to find 7ⁿ – 3ⁿ is divisible by which number

Consider P (n): 7n – 3n

P (1): 7¹ − 3¹ = 4

Thus, 7n – 3n is divisible by 4

Let P (k) is true for n = K

⇒ 7^k − 3^k is divisible by 4

So, 7n – 3n = 4d

Now, prove that P (k+1) is true.

⇒ 7^(k+1) − 3^(k+1) = 7^(k+1) −7.3^k + 7.3^k − 3^(k+1)

= 7(7^k − 3^k) + (7 − 3)3^k

= 7(4d) + (7 − 3)3^k

= 7(4d) + 4.3^k

= 4(7d + 3^k)​

Hence, P (n): 7ⁿ − 3ⁿ is divisible by 4 is true.

Concept:

Principle of mathematical induction: Let P(n) be the statement. 

P(n) is true for n = 1. Let P(n) is true for n = k then if P(n) is true for n = k + 1 then P(n) is true for all natural numbers n.

Also, for n ≥ 3, n² > 2n + 1 [This can be proved using the Principle of mathematical induction.]

Explanation: 

Given Statement: n2 n

• For n = 1: 1² 1, which is true.
• For n = 2: 2² 2, which is not true.
• For n = 3: 3² 3, which is not true.
• For " id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">
  n = 4: 4² 4, which is not true.
• For " id="MathJax-Element-10-Frame" role="presentation" style="position: relative;" tabindex="0">
  n = 5: 5² 5, which is true.
• For n = 6: 62 6, which is true.

Now, let's prove it for all natural numbers greater than 5.

Base Case (n = 5):

Which is true.

Let's assume for some arbitrary k ≥ 5, it holds that .

We need to show that .

Multiply both sides by 2, we get

⇒ 

As  2k + 1\)

⇒ 

⇒ 

Which is the required result.

Thus, Given inequality holds true for all natural numbers greater than or equal to 5.

Thus,  Option 1 is the correct answer.

Concept:

Principle of mathematical induction: Let P(n) be the statement. 

P(n) is true for n = 1. Let P(n) is true for n = k then if P(n) is true for n = k + 1 then P(n) is true for all natural numbers n.

Also, for n ≥ 3, n² > 2n + 1 [This can be proved using the Principle of mathematical induction.]

Explanation: 

Given Statement: n2 n

• For n = 1: 1² 1, which is true.
• For n = 2: 2² 2, which is not true.
• For n = 3: 3² 3, which is not true.
• For " id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">
  n = 4: 4² 4, which is not true.
• For " id="MathJax-Element-10-Frame" role="presentation" style="position: relative;" tabindex="0">
  n = 5: 5² 5, which is true.
• For n = 6: 62 6, which is true.

Now, let's prove it for all natural numbers greater than 5.

Base Case (n = 5):

Which is true.

Let's assume for some arbitrary k ≥ 5, it holds that .

We need to show that .

Multiply both sides by 2, we get

⇒ 

As  2k + 1\)

⇒ 

⇒ 

Which is the required result.

Thus, Given inequality holds true for all natural numbers greater than or equal to 5.

Thus,  Option 1 is the correct answer.

Calculation

Let the statement P(n) be defined as

 is a natural number for all n ∈ N.

Step 1 : For n = 1, P(1) :  = 1 ∈ N

Hence, it is true for n = 1.

Step II : Let it is true for n = k,

i.e. 

Step III: For n = k + 1

=  + 

=  + 

= λ + k⁴ + 2k³ + 3k² + 2k + 1

[using equation (i)]

which is a natural number, since λk ∈ N. Therefore, P(k + 1) is true, when P(k) is true, Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.

Hence option 4 is correct

We have

Hence is divisible by 225.

Concepts:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P (1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P (k) implies the truth of P (k + 1).

Then, P (n) is true for all natural numbers n

Calculation:

Given:

P (n) = 121ⁿ – 25ⁿ + 1900ⁿ – (-4) ⁿ

Now, P (1) = 121¹ – 25¹ + 1900¹ – (-4)¹

⇒ P (1) = 121 – 25 + 1900 + 4

⇒ P (1) = 2000

Therefore we can say that P (n) is divisible by 2000.

Calculation:

S = 72n + 16n - 1

For n = 1

S = 49 + 16 - 1 = 64

For n = 2 

S = 2401 + 32 - 1 = 2432 = 64 × 32, which is divisible by 64

∴ Option 1 is correct

Concept:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P(1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P (k) implies the truth of P (k + 1).

Then, P (n) is true for all natural numbers n

Calculation:

We have to find 7ⁿ – 3ⁿ is divisible by which number

Consider P (n): 7n – 3n

P (1): 7¹ − 3¹ = 4

Thus, 7n – 3n is divisible by 4

Let P (k) is true for n = K

⇒ 7^k − 3^k is divisible by 4

So, 7n – 3n = 4d

Now, prove that P (k+1) is true.

⇒ 7^(k+1) − 3^(k+1) = 7^(k+1) −7.3^k + 7.3^k − 3^(k+1)

= 7(7^k − 3^k) + (7 − 3)3^k

= 7(4d) + (7 − 3)3^k

= 7(4d) + 4.3^k

= 4(7d + 3^k)​

Hence, P (n): 7ⁿ − 3ⁿ is divisible by 4 is true.

Formula used:

1. (x + y)ⁿ = ⁿC₀ (xⁿ) (y⁰) + ⁿC₁ (x^n-1)y + ......ⁿC_n (x⁰)(yⁿ)

2. ⁿC_r = ⁿC_n-r

3. nC0 = 1

4. ⁿC₁ =  n

Calculations:

Let y = 4n - 3n - 1 

⇒ y = (1 + 3)n - 3n - 1 

By using the above formula

⇒ y = nC0 (1n) (30) + nC1 (1)n-1(3¹) + ......nCn (10)(3n) - 3n - 1

Again, using the formula (1), (2) & (3)

⇒ y = nC0 + nC1 (3) + nC₂ (3²) + ......nC_n (3n) - 3n - 1

⇒ y = 1 + 3n + ⁿC₂ 3² + .....+ nCn3ⁿ - 3n - 1

⇒ y = 3²(nC2 32 + ⁿC₃3³ .....+ nCn3n) 

⇒ y = 9 × Integer

∴ 4n - 3n - 1 is always divisible by 3 & 9 both.

Alternate Method

Since, n is positive, put n = 1, 2, 3....

Put n = 1

⇒ 41 - 3(1) - 1 = 4 - 3 - 1 = 0

Which is divisible by 3, 9, 8, 27

Put n = 2,

⇒ 42 - 3(2) - 1 = 16 - 6 - 1 = 9

Which is divisible by 3, 9

Put n = 3

⇒ 43 - 3(3) - 1 = 64 - 9 - 1 = 54

Which is divisible by 3, 9, 27

Put n = 4, 

⇒ 44 - 3(4) - 1 = 256 - 12 - 1 = 243

Which is divisible by 3, 9, 27

Put n = 5, 

⇒ 45 - 3(5) - 1 = 1024 - 15 - 1 = 1008 ,

Which is divisible by 3, 9, 27

Continued so on....

∴ 4n - 3n - 1 is always divisible by 3 & 9 both

Note: This is BSF RO Official Paper (Conducted on 22-Sep-2019) official paper and there were options 3, 9, 8 & 27 in the option. Since the relation is satisfied for both 3 & 9 therefore, we have changed the option. 

Concepts:

Principle of Mathematical Induction:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P (1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., the truth of P (k) implies the truth of P (k + 1). 

Then, P (n) is true for all natural numbers n

Calculation:

Given

P(n) = 2 + 4 + ......+ 2n

Put n = 1

P(1) = 2

Hence, P(n) = n(n + 1) + 2 is not true for n = 1 

So, The Principle of Mathematical Induction is not applicable and nothing can be said about the validity of the statement P(n) = n(n + 1) + 2

Concept:

Let us suppose, P(n) = 7n - 2n .

If n =1, then P(1) = 5

⇒ For, n = 1 we can conclude that P(n) is divisible by 5

Let us assume, for some positive integer k, P(k) = 7k - 2k is divisible by 5

⇒ 7k - 2k = 5 × d where d ∈ N.

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

P(k + 1) = 7(k + 1) - 2(k + 1) 

⇒ P(k + 1) = 7(k + 1) - 3(k + 1) = 7(k + 1) - 7 × 2k + 7 × 2k - 2(k + 1) 

⇒ P(k + 1) = 7 × (7k - 2k) + 2k × (7 - 2)

We know that P(k) is true i.e 7k - 2k = 5 × d where d ∈ N

⇒ P(k + 1) = 7 × (5d) + 2k × 5

⇒ P(k + 1) = 5 × (7d + 2k)

⇒ P(k + 1) = 5 × q where q = (7d + 2k)

Hence, P(k + 1) is also divisible by 5.

So, P(n) = 7n - 2n  is divisible by 5 for all positive integers n.

Shortcut Trick

P(n) = 7ⁿ – 2ⁿ

Put n = 1

7n – 2ⁿ = 7¹ – 2¹ = 7 – 2 = 5

which is divisible by 5

Put n = 2

7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45 (divisible by 5)

which is 

Put n = 3

7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335 (divisible by 5)

Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5

Concept:

• Mathematical induction: It is a technique of proving a statement, theorem, or formula which is assumed to be true, for every natural number n.
• By generalizing this in the form of a principle that we would use to prove any mathematical statement is called the principle of mathematical induction.

Calculations:

Consider

Clearly, the r^th term from the above series is 

Let the r^th term be    ....(1)

now multiply and divide by 2 in (1)

⇒ 

⇒ 

add and subtract r in the numerator

⇒ 

⇒ 

⇒    (2)

Now put r = 1 in (2), then we have

put r = 2 in (2)

⇒      

put r = 3 in (2)

⇒ 

      .  .  .  .  .  .  . 

put r = n in (2) 

Now, add all these equations and we get

Hence, the correct answer is option 4).

Concepts:

Principle of Mathematical Induction:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P (1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., the truth of P (k) implies the truth of P (k + 1). 

Then, P (n) is true for all natural numbers n

Calculation:

Given:

T(k) = 1 + 3 + 5 + ........ + (2k - 1) = k2 + 10

Put k = 1

T(1) = 2 × 1 - 1 = 1² + 10

1 ≠ 10, LHS ≠ RHS

∴ T(1) is not true

Let T(k) is true

1 + 3 + 5 + ......... + (2k - 1) = k2 + 10     ---(1)

OR, 1 + 3 + 5 + ....... + (2k - 1) + (2k + 1) = k2 + 10 + 2k + 1  Using eqn (1)

⇒ 1 + 3 + 5 + ........ + (2k - 1) + (2k + 1) = (k + 1)² + 10

∴ T(k + 1) is true

i.e T(k) is true ⇒  T(k + 1) is true (Option Two is correct)

T(n) is not true for all n ϵ N, as T(1) is not true.

Concepts:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P (1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P (k) implies the truth of P (k + 1).

Then, P (n) is true for all natural numbers n

Calculation:

Given: 

P(n) = 

Take n = 1

P(1) = 

Therefore we can say that P (n) is divisible by 11

Concept:

Calculation:

Given:

P(n): 3n 

This can be solved directly by hit and trial method, putting the option in expression and checking its validity

We will choose first the smallest number from the options

Putting n = 3

P(n) = 3n 3 

⇒ 27  3 × 2 × 1, 27  6 , Hence Option 2 is wrong

Putting n = 6

P(n) = 3n 3 

P(n) = 3n 6 

1029  720 Hence Option 3 is wrong

Put n = 7

P(n) = 3n 7 

2187