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

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നേടുക Principles of Mathematical Induction ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Principles of Mathematical Induction MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Principles of Mathematical Induction MCQ Objective Questions

Top Principles of Mathematical Induction MCQ Objective Questions

Principles of Mathematical Induction Question 1:

Rahul was asked to prove a statement P(n) by the principle of mathematical induction. He proved that P(k + 1) is true whenever P(k) is true for all natural numbers k and also that P(9) is true. Then P(n) is true:

  1. for all n > 8
  2. for all n < 9
  3. for all natural numbers n
  4. for all n ≥ 9

Answer (Detailed Solution Below)

Option 4 : for all n ≥ 9

Principles of Mathematical Induction Question 1 Detailed Solution

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

Principles of Mathematical Induction Question 2:

Using the principle of mathematical induction, find the value of \(\frac{1}{{1.2.3}}\; + \;\frac{1}{{2.3.4}}\; + \; \ldots \frac{1}{{\left\{ {n\left( {n\; + \;1} \right)\left( {n\; + \;2} \right)} \right\}}}\)

  1.  \(\frac{n (n+1)}{4(n +2)(n +3)}\)
  2.  \(\frac{n (n+3)}{4(n +1)(n +2)}\)
  3.  \(\frac{n (n+2)}{4(n +1)(n +3)}\)
  4. None of the above

Answer (Detailed Solution Below)

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

Principles of Mathematical Induction Question 2 Detailed Solution

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

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

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

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

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

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

Principles of Mathematical Induction Question 3:

If n ϵ N, \((\frac{n~+~1}{2})^n\geq n!\) is true when 

  1. n ≥ 1
  2. n ≥ 2
  3. n > 1
  4. n > 2

Answer (Detailed Solution Below)

Option 1 : n ≥ 1

Principles of Mathematical Induction Question 3 Detailed Solution

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

Principles of Mathematical Induction Question 4:

For every natural number k, which of the following is true?

  1. (mn)= mknk
  2. mk2 = n + 1
  3. (m+n)k = k + 1
  4. mkn = mnk

Answer (Detailed Solution Below)

Option 1 : (mn)= mknk

Principles of Mathematical Induction Question 4 Detailed Solution

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 = mpnp

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.

Principles of Mathematical Induction Question 5:

If 22n - 1 is divisible by k for all n ∈ N, then the value of k is:

  1. 6
  2. 3
  3. 7
  4. 2

Answer (Detailed Solution Below)

Option 2 : 3

Principles of Mathematical Induction Question 5 Detailed Solution

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.
  • Na divided by d will give a remainder of ra.

 

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.

Principles of Mathematical Induction Question 6:

Which of the following steps is mandatory in the principle of mathematical induction?

  1. Inductive reference
  2. Inductive hypothesis
  3. Minimal set representation
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : Inductive hypothesis

Principles of Mathematical Induction Question 6 Detailed Solution

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.

Principles of Mathematical Induction Question 7:

For all n ϵ N, \((1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2n~+~1)}{n^2}))\) is equal to

  1. \(\frac{(n~+~1)^2}{2}\)
  2. \(\frac{(n~+~1)^3}{3}\)
  3. \((n+1)^2\)
  4. None of these
  5. Not Attempted

Answer (Detailed Solution Below)

Option 3 : \((n+1)^2\)

Principles of Mathematical Induction Question 7 Detailed Solution

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

Principles of Mathematical Induction Question 8:

P(n): 1.1! + 2.2! + 3.3! + n.n! = (n + 1)! – 1, then P(n) statement is true-

  1. For n > 1
  2. For n > 4
  3. For all values of n 
  4. For all negative values of n

Answer (Detailed Solution Below)

Option 3 : For all values of n 

Principles of Mathematical Induction Question 8 Detailed Solution

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.

Principles of Mathematical Induction Question 9:

For all n ≥ 2, \({n^2}\left( {{n^4} - 1} \right)\) is divisible by ________

  1. 60
  2. 50
  3. 40
  4. 70

Answer (Detailed Solution Below)

Option 1 : 60

Principles of Mathematical Induction Question 9 Detailed Solution

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

From the options, these are divisible by 60.

Principles of Mathematical Induction Question 10:

For every positive integer n, 7n – 3n is divisible by

  1. 2
  2. 4
  3. 5
  4. 6
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 4

Principles of Mathematical Induction Question 10 Detailed Solution

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− 3k 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) −7.3k + 7.3k − 3(k+1)

= 7(7− 3k) + (7 − 3)3k

= 7(4d) + (7 − 3)3k

= 7(4d) + 4.3k

= 4(7d + 3k)​

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