Principles of Mathematical Induction MCQ Quiz in தமிழ் - Objective Question with Answer for Principles of Mathematical Induction - இலவச PDF ஐப் பதிவிறக்கவும்

Concepts:

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

1. The statement is true for n = 1, i.e., P (1) is true, and

2. 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:

For, k = 1

Option 1,

(mn)¹ = m¹n¹, mn = mn (LHS = RHS) hence it is true

Let us assume that the statement is true for, k = p

(mn)^p = m^pn^p

Multiplying the above equation with mn, we get mn we get,

(mn)p + 1 = mp +1 np + 1

mp + 1 np + 1 = mp + 1 np + 1

Hence the given expression is true for every natural number k.

Option 2, 3 and 4 does not satisfy for k = 1, Hence 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:

Given: P (n) is true

For n = 1

P (1): 3¹ > 1¹

⇒ 3 > 1

∴ It is true for n = 1

It is given that P (k) is true, so P (k) is true for n = k

P (K) ⇒ 3^k > k^k

Now,

P (k +1) = 3^k+1 > (k + 1) ^k+1

3^k.3¹ > (k + 1) ^k (k + 1)

By the principle for mathematics production,

P (n + 1) is true, when P (n) is true 

Explanation

For a statement P(n) to be true for all \( n \in \mathbb{N}\) 

The principle of mathematical induction requires two conditions:

1. Base Case: Verify that the statement is true for the initial value, often ( n = 1 ).
2. Inductive Step: Assume the statement is true for some arbitrary natural number k, i.e., P(k) is true. Then prove that P(k+1) is also true based on this assumption.

Based on this understanding, among the given options, the condition that must be satisfied is: P(1) must be true

Thus, the correct answer is: 3) P(1) must be true

Calculation:

If P(a) is true and the truth of P(n) implies that P(n+1) is also true, then we can conclude that P(n) is true for all n ≥ a, where n ∈ N.

Here P(9) is true.

∴ We can conclude that P(n) is true for all n ≥ 9

Concept:

Principle of Mathematical Induction: Let P(n) be the statement. 

Base Case: P(n) is true for n = 1.

Inductive Hypothesis: Let P(n) is true for n = k then

Inductive Step: if P(n) is true for n = k + 1.

⇒  P(n) is true for all natural number n.

Thus, Option 2 is the correct answer.

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:

Let P(n) be defined as

\(P(n)=(1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2n~+~1)}{n^2}))=(n+1)^2\)

Put n = 1

P(1) = \((1+\frac{3}{1})\) = (1 + 1)² 

4 = 4 P(1) is true 

Let it is true for n = k

\(P(k)=(1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2k~+~1)}{k^2}))=(k+1)^2\) ....(1)

for n = k + 1

\(P(k+1)=[(1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2k~+~1)}{k^2}))=(k+1)^2](1~+~\frac{2k+1+2}{(k+1)^2})=(k+1)^2(1+\frac{2k+3}{(k+1)^2})\)  Using Equation (1)

= \((k +1)^2[\frac{(k+1)^2+2k+3}{(k+1)}]\)

= k² + 2k + 1 + 2k + 3

= (k +2)² = [(k+1) + 1]²

Therefore, P(k +1) is true, Hence From the Principle of Mathematical Induction, the statement is true for all natural numbers n

Concept:

If a number N is divided by d to give quotient q and remainder r, then:

• N = dq + r = d(q + 1) + (r - d) = d(q - 1) + (r + d).

  It means that a remainder of r is equivalent to a remainder of r - d which is also equivalent to a remainder of r + d, where d is the divisor.

• N^a divided by d will give a remainder of r^a.

Calculation:

Since, 22n - 1 is 1 less than 22n, let's check for the divisors which definitely give a remainder of 1 on dividing 22n.

2 gives a definite remainder of 2 - 3 = -1 only when divided by 3.

∴ 22 gives a definite remainder of (-1)² = 1 on division by 3.

⇒ 22n gives a definite remainder of 1ⁿ = 1 on division by 3.

⇒ 22n - 1 gives a definite remainder of 1 - 1 = 0 on division by 3.

⇒ 22n - 1 is always divisible by 3.

∴ The value of k is 3.

For every positive integers n ≥ 2, \(f\left( n \right) = {n^2}\left( {{n^4} - 1} \right)\)

For n = 2, \(f\left( 2 \right) = {2^2}\left( {{2^4} - 1} \right) = 60\)

For n = 3, \(f\left( 2 \right) = {3^2}\left( {{3^4} - 1} \right) = 720\)

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)​

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 < 2n

• For n = 1: 1² < 2¹, which is true.
• For n = 2: 2² < 2², which is not true.
• For n = 3: 3² < 2³, which is not true.
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  n = 4: 4² < 2⁴, which is not true.
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  n = 5: 5² < 2⁵, which is true.
• For n = 6: 62 < 26, which is true.

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

Base Case (n = 5):

\(5^2 = 25 < 2^5 = 32\)

Which is true.

Let's assume for some arbitrary k ≥ 5, it holds that \( k^2 < 2^k\).

We need to show that \((k+1)^2 < 2^{k+1}\).

\( k^2 < 2^k\)

Multiply both sides by 2, we get

\(2k^2 < 2^{k+1}\)

⇒ \(k^2 + k^2 < 2^{k+1}\)

As \(k^2 > 2k + 1\)

⇒ \(k^2 + 2k + 1 < 2^{k+1}\)

⇒ \((k+ 1)^2 < 2^{k+1}\)

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.