Principles of Mathematical Induction MCQ Quiz in मल्याळम - Objective Question with Answer for Principles of Mathematical Induction - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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.

Calculation

Let us suppose, P(n) = 5n - 5.

If n =1, then P(1) = 0

⇒ For, n = 1 we can conclude that P(n) is divisible by 4.

Let us assume, for some positive integer k, P(k) = 5^k - 5 is divisible by 4.

⇒ 5² - 5 = 4 × d where d ∈ N.

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

P(k + 1) = 5^k+1 - 5

⇒ P(k + 1) = 5k+1 - 5 =5.5k - 5 =  5.5k - 25 + 20 

⇒ P(k + 1) = 5(5k - 5) + 20 

We know that P(k) is true i.e 52 - 5 = 4 × d where d ∈ N

⇒ P(k + 1) = 5 × (4d) + 20

⇒ P(k + 1) = 4 × (5d + 5)

⇒ P(k + 1) = 4 × q where q = (5d + 5)

Hence, P(k + 1) is also divisible by 4.

∴ Option 2 is correct

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

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{n(n+1)(n+2)}\)

Clearly, the rth term from the above series is 

Let the rth term be \(u_r=\frac{1}{r(r+1)(r+2)}\)   ....(1)

now multiply and divide by 2 in (1)

⇒ \(u_r=\frac{1\times 2}{2r(r+1)(r+2)}\)

⇒ \(u_r=\frac{2}{2r(r+1)(r+2)}\)

add and subtract r in the numerator

⇒ \(u_r=\frac{(r+2) - r}{2r(r+1)(r+2)}\)

⇒ \(u_r=\frac{1}{2}\big[\frac{(r+2) }{r(r+1)(r+2)}- \frac{r }{r(r+1)(r+2)}\big]\)

⇒ \(u_r=\frac{1}{2}\big[\frac{1}{r(r+1)}- \frac{1}{(r+1)(r+2)}\big]\)   (2)

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

\(u_1=\frac{1}{2}\big[\frac{1}{1.2}- \frac{1}{2.3}\big]\)     

put r = 2 in (2)

⇒ \(u_2=\frac{1}{2}\big[\frac{1}{2.3}- \frac{1}{3.4}\big]\)     

put r = 3 in (2)

⇒ \(u_3=\frac{1}{2}\big[\frac{1}{3.4}- \frac{1}{4.5}\big]\)

      .  .  .  .  .  .  . 

put r = n in (2) 

\(u_r=\frac{1}{2}\big[\frac{1}{n(n+1)}- \frac{1}{(n+1)(n+2)}\big]\)        

Now, add all these equations and we get

\(S_n=\displaystyle\sum_{r=1}^{n} u_r=\frac{1}{2}\bigg[\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\bigg ]\)

\(=\frac{1}{4(n+1)(n+2)}\big[(n+1)(n+2)-2\big ]\)

\(=\frac{1}{4(n+1)(n+2)}\big[(n^2+n+2n+2-2\big ]\)

\(=\frac{1}{4(n+1)(n+2)}\big[n^2+3n\big ]\)

\(=\frac{n(n+3)}{4(n+1)(n+2)}\)

Concept:

\(\rm n! = n×(n-1)×(n -2)....× 3×2×1\)

Calculation:

Given:

\(P(n)=(\frac{n~+~1}{2})^n≥ n!\)

Put n = 1

\(P(1) =(\frac{1~+~1}{2})^1≥ 1!\) , 1 ≥ 1

Put n = 2

\(P(2)=(\frac{2~+~1}{2})^2≥ 2!~=(\frac{3}{2})^2≥ 2 × 1\)

2.25 ≥ 2

Put n = 3

\(P(3) =(\frac{3~+~1}{2})^3≥ 3!\)

8 ≥ 3 × 2 × 1 , 8 ≥ 6

Hence, The given expression P(n) is true for n ≥ 1

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:

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.

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

Proof:

To prove \(P(n): 1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + … + n ⋅ n! = (n + 1)! - 1\) for all positive integers n, we can use mathematical induction.

Base Case: Putting n = 1  into the equation, we get:

1 ⋅ 1! = (1 + 1)! - 1

⇒ 1 ⋅ 1 = (2)! - 1

⇒ 1 = 2 - 1

⇒ 1 = 1

Thus, the Base step is true.
   
Inductive Hypothesis: Assume that the statement  P(k)  is true for some arbitrary positive integer k, i.e., 

\(1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + … + k ⋅ k! = (k + 1)! - 1 \) . . . (i)

Inductive Step: We want to prove that if  P(k)  is true, then  P(k + 1)  is also true.

Adding  (k + 1) ⋅ (k + 1)!  to both sides of the equation (i), we get:
   
1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + … + k ⋅ k! + (k + 1) ⋅ (k + 1)! = (k + 1)! - 1 + (k + 1) ⋅ (k + 1)! 

RHS = (k + 1)! ⋅ (1 + k + 1) - 1 

⇒ RHS = (k + 1)! ⋅ (k + 2) - 1 

⇒ RHS  = (k + 2)! - 1 

Therefore,  P(k + 1)  is true.

Since  P(1)  is true and  P(k)  being true implies  P(k + 1)  is true, by mathematical induction,  P(n)  is true for all positive integers n.

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)​