NCERT Exemplar: Principle of Mathematical Induction

# NCERT Exemplar: Principle of Mathematical Induction | Mathematics (Maths) Class 11 - Commerce PDF Download

Q.1. Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.
Ans.
The required statement is P(n) = 2n < n!
Justification: P(n) : 2n < n!
P(1) : 2.1 < 1!
⇒ 2 < 1 not true
P(2) : 2.2 < 2!
⇒ 4 < 2.1
⇒ 4 < 2 not true
P(3) : 2.3 < 3!
⇒ 6 < 3.2.1
⇒ 6 < 6 not true
P(4) : 2.4 < 4!
⇒ 8 < 4.3.2.1
⇒ 8 < 24 True
P(5) : 2.5 < 5!
⇒ 10 < 5.4.3.2.1
⇒ 10 < 120 True
Hence, P(n) = 2n < n! is not true for P(1), P(2) and P(3) but it is true for all values of n ≥ 4.

Q.2. Give an example of a statement P(n) which is true for all n. Justify your answer.
Prove each of the statements in Exercises 3 - 16 by the Principle of Mathematical Induction :
Ans.
The required statement is
P(n) : 1 + 2 + 3 + ... + n =
Justification: P(1) :
P(k) : 1 + 2 + 3 + ... + k = Let it be true.
P(k + 1) : 1 + 2 + 3 + ... + k + (k + 1)

Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have p(n) is true for all n.

Q.3.  4n – 1 is divisible by 3, for each natural number n.
Ans.
Let P(n) : 4n – 1
Step 1: P(1) = 4 – 1 = 3 which is divisible by 3, so it is true.
Step 2: P(2) = 4k – 1 = 3λ . Let it be true.
Step 3: P(k + 1)  =  4k + 1 – 1
= 4k .4 – 1 = 4.4k   4 + 3 = 4(4k  1) + 3
= 4 (3λ)  +  3 (from Step 2)
= 3[4λ + 1] which is true as it is divisible by 3.
Hence, P(k + 1) is true whenever P(k) is true.

Q.4. 23n – 1 is divisible by 7, for all natural numbers n.
Ans.
Let P(n) : 23n – 1
Step 1: P(1) = 23.1 1 = 8  1 = 7 which is divisible by 7.
So, P(1) is true.
Step 2: P(k) = 23k – 1 = 7l. Let it be true.
Step 3: P(k + 1) =  23(k + 1) – 1
= 23k + 3 – 1 = 23.23k   8 + 7 = 8.23k  8 + 7
= 8(23k  1) + 7  (from Step 2)
= 8.7λ + 7
= 7(8λ + 1) which is true as it is divisible by 7
Hence, P(k + 1) is true whenever P(k) is true.

Q.5. n3 – 7n + 3 is divisible by 3, for all natural numbers n.
Ans.
Let P(n) : n3  7n + 3
Step 1: P(1) : (1)3  7(1) + 3
= 1  7 + 3 = - 3  which is divisible by 3.
So, P(1) is true    .
Step 2: Assume P(k) is true for some  K∈N P(k) : k3  7k + 3 = 3λ, λ∈N.
k3 =  + 7k  3..............(i)
Step 3: Now we have to prove P(k + 1) : (k + 1)3  7(k + 1) + 3 is divisible by 3.
Now P(k + 1) : (k + 1)3  7(k + 1) + 3= k3 + 1 + 3k2 + 3k  7k  7 + 3
= k3 + 3k2  4k  3
= (3λ + 7k  3) + 3k2  4k – 3 (using equation (i))
= 3k2 + 3k + 3λ  6
= 3(k2 + k + λ  2) is divisible by 3.
⇒ P(k + 1) is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have p(n) is true for all n.

Q.6. 32n – 1 is divisible by 8, for all natural numbers n.
Ans.
Let P(n) : 32n  1
Step 1: P(1) : 32  1 = 9  1 = 8 which is divisible by 8.
So, P(1) is true.
Step 2: Assume P(k) is true for some P(k) : 32k  1 = 8λ, λ∈N.     Step 3: Now we have to prove P(k + 1) : 32(k + 1)  1  is divisible by 8.
P(k + 1) : 32(k + 1)  1
= 32k + 2 1 = 32.32k   9 + 8 = 9(32k   1) + 8
= 9.8λ + 8 (from Step 2)
= 8[9λ + 1] is divisible by 8.
⇒ P(k + 1) is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have p(n) is true for all n.

Q.7. For any natural number n, 7n – 2n is divisible by 5.
Ans.
Let P(n) : 7n – 2n
Step 1: P(1) : 71 – 21 = 5 which is divisible by 5.
So it is true for P(1).
Step 2: P(k) : 7k – 2k = 5λ. Let it be true for P(k).
Step 3: P(k +1) = 7k + 1 – 2k + 1
= 7k + 1 + 7k.2 – 7k.2 – 2k + 1
= (7k + 1 – 7k.2) + 7k.2 – 2k + 1)
= 7k(7 – 2) + 2.(7k – 2k)
= 5.7k + 2.5λ (from Step 2)
= 5(7k + 2λ) which is divisible by 5.
So, it is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

Q.8. For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
Ans.
Let P(n) : xn – yn
Step 1: P(1) : x1 – y1 = x – y which is divisible by x – y.
So P(1) is true.
Step 2: P(k) : xk – yk = (x – y)λ. Let it be true.
Step 3: P(k + 1) = xk + 1 – yk + 1 = xk + 1 – xky – xky – yk + 1
= (xk + 1 – xky) + (xky – yk + 1)
= xk(x – y) + y(xk – yk)
= xk (x – y) + y.(x – y)λ (from Step 2)
= (x – y) (xk + yλ)which is divisible by (x – y).
So, it is true for P(k + 1).

Q.9. n3 – n is divisible by 6, for each natural number n ≥ 2.
Ans.
Let P(n) : n3 – n
Step 1: P(2) : 23 – 2 = 6 which is divisible by 6. So it is true for P(2).
Step 2: P(k) : k3 – k = 6λ. Let it be true for k ≥ 2
⇒ k3 = 6λ + k ...(i)
Step 3: P(k + 1)
= (k + 1)3 – (k + 1)
= k3 + 1 + 3k2 + 3k – k – 1
= k3 – k + 3(k2 + k)
= 6λ + 3(k2 + k) [from (i)]
We know that 3(k2 + k) is divisible by 6 for every value of k ∈ N.
Hence P(k + 1) is true whenever P(k) is true.

Q.10. n (n+ 5) is divisible by 6, for each natural number n.
Ans.
Let P(n) : n(n2 + 5)
Step 1: P(1) : 1(1 + 5) = 6 which is divisible by 6. So it is true for P(1).
Step 2: P(k) : k(k2 + 5) = 6λ. Let it be true
⇒ k3 + 5k =
⇒ k3 = 6λ – 5k ...(i)
Step 3: P(k + 1) = (k + 1)[(k + 1)2 + 5]
= (k + 1)[k2 + 1 + 2k + 5]
= (k + 1)[k2 + 2k + 6]
= k3 + 2k2 + 6k + k2 + 2k + 6
= k3 + 3k2 + 8k + 6
= k3 + 5k + 3k2 + 3k + 6
= 6λ – 5k + 5k + 3(k2 + k + 2)  [From (i)]
=  + 3(k2 + k + 2)
We know that k2 + k + 2 is divisible by 2 for each value of k ∈ N,
so, let k2 + k + 2 = 2m.
So P(k + 1)  =   +  3.2m = 6(λ + m) which is divisible by 6.
Hence, P(k + 1) is  true whenever P(k) is true.

Q.11. n2 < 2n for all natural numbers n ≥ 5.
Ans.
Let P(n) : n2 < 2n for all natural number, n  5.
Step 1: P(5) : 52 < 25  25 < 32 which is true. So P(5) is true.
Step 2: Assume P(k) is true for some k∈N, k  5 ⇒ P(k) : k2 < 2k.............(i) is true.
Step 3: Now we have to prove P(k + 1) : (k + 1)2 < 2k + 1 is true where k  5.
Now we have (k + 1)2  = k2 + 2k + 1 < 2k + 2k + 1 .............(ii) [using equation(i)]
Now let 2k + 2k + 1 < 2k + 1
⇒ 2k + 2k + 1 < 2k . 2 =  2k + 2k
⇒ (2k + 1) < 2k  which is true for K∈N , k  5 ….........(iii)
So (k + 1)2 < 2k + 2k   [using (ii) and (iii)]
⇒ (k + 1)2 < 2.2k
⇒(k + 1)2 < 2k + 1.
⇒ P(k + 1) is true.
Hence, P(k + 1) is true whenever P(k) is  true.
Therefore by the principle of mathematical induction we have p(n) is true for all n  5.

Q.12. 2n < (n + 2)! for all natural number n
Ans.
Let p(n) : 2n < (n + 2)! for all natural number n.
Step 1: P(1): 2.1 < (1 + 2)!
⇒ 2 < 3! ⇒ 2 < 6 which is true  (∴ 3! = 3 × 2 × 1 = 6)
So P(1) is true.
Step 2: Assume P(k) is true for some K∈N ⇒ P(k): 2k < (k + 2)! is true..
Step 3: Now we have to prove P(k + 1) : 2(k + 1) <  (k + 1 + 2)! is true.
Since 2k < (k + 2)! (from Step 2)
⇒ 2k + 2 < (k + 2)! + 2
2(k + 1) < (k + 2)! + 2............(i)
Now let  (k + 2)! + 2 < (k + 3)!............(ii)
⇒ 2 < (k + 3)! - (k + 2)!
⇒2 <  (k + 2)! [k + 3-1]
2 < (k + 2)!.(k + 2) which is true for any natural number k.
2(k + 1) < (k + 3)!   [using (i) and (ii)]
⇒ 2(k + 1) < (k + 2 + 1)!
⇒ P(k + 1) is true.
Hence, P(k + 1) is true whenever P(k) is true.     Therefore by the principle of mathematical induction we have p(n) is true for all n.

Q.13. for all natural numbers n ≥ 2.
Ans.
Let P(n)
Step 1: P(2) :which is true.
Step 2: P(k)     Let it be true.
Step 3: P(k + 1) :
From Step 2, we have

Now if

...(ii)
From eqn. (i) and (ii) we get

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

Q.14. 2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.
Ans.
Let P(n) :  2 + 4 + 6 + ... + 2n = n2 + n, ∀n ∈ N
Step 1: P(1) : 2 = 12 + 1 = 2
which is true for P(1)
Step 2: P(k) : 2 + 4 + 6 + ...+ 2k = k2 + k. Let it be true.
Step 3: P(k + 1) : 2 +  4 + 6 +...+ 2k + 2k + 2
= k2 + k + 2k + 2
= k2 + 3k + 2
= k2 + 2k + k + 1 + 1
= (k + 1)2 + (k + 1)
Which is true for P(k + 1)
So, P(k + 1) is true whenever P(k) is true.

Q.15. 1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.
Ans.
Let P(n) : 1 + 2 +  22 +  + 2n = 2n + 1  1, n∈N.
P(n) : 20 + 21 + 22 +  + 2n = 2n+1  1
Step 1: P(1) L.H.S = 20 +21 = 1 + 2 = 3.
R.H.S = 21 + 1  1 = 22  1 = 4 – 1 = 3 .
L.H.S = R.H.S. So P(1) is true.
Step 2: Assume P(k) is true for some k∈N
⇒ P(k) : 20 + 21 + 22 +  + 2k = 2k + 1  1..........(i)
Step 3: Now we have to prove P(k + 1) : 20 + 21 + 22 +…+ 2k + 2k + 1 = 2(k + 1) + 1 1.
Adding 2k + 1 on both sides of equation (i)
20 + 21 + 22 +  + 2k + 2k + 1
= 2k + 1  1 + 2k + 1 = 2.2k + 1  1
= 2k + 2  1
= 2(k + 1) + 1  1
⇒ P(k + 1) is true.

Q.16. 1 + 5 + 9 + ... + (4n – 3) = n (2n – 1) for all natural numbers n.
Ans.
Let P(n) : 1 +  5  +  9 +  + (4n  3) = n(2n  1),  n ∈ N
Step 1: P(1) : 1 = 1(2.1  1) = 1 which is true for P(1)
Step 2: P(k) : 1 + 5 + 9 +...+ (4k – 3) = k(2k – 1). Let it be true.
Step 3: P(k + 1) : 1 + 5 + 9 +...+ (4k  3) + (4k + 1)
= k(2k  1) + (4k + 1) = 2k2 – k + 4k + 1
= 2k2 + 3k + 1 = 2k2 + 2k + k + 1
= 2k(k + 1) + 1(k + 1) = (2k + 1)(k + 1)
= (k + 1)(2k + 2  1) = (k + 1) [2(k + 1)  1]
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

Q.17. A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak–1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.
Ans.
Given that:
a1 = 3
a2 = 7a2 – 1 = 7.a1 = 7.3 = 21
a3 = 7.a3 – 1 = 7.a2 = 7.21 = 147...
Let P(n) : an = 3.7n  1, ∀n∈N
Step 1: P(2): a2 = 3.72  1 = 21  21. So P(2) is true.
Step 2: P(k): ak = 3.7k – 1. Let it be true.
Step 3: ak = 7ak – 1 (given)
Put k = k + 1
ak + 1 = 7ak = 7(3.7k – 1) = 3.7k + 1  1 = 3.7(k + 1)  1
which is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Q.18. A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
Ans.
We have b0 = 5 and bk = 4 + bk – 1
⇒b0 = 5, b1 = 4 + b0 = 4 + 5 = 9 and b2 = 4 + b1 = 4 + 9 = 13
Let P(n) : bn = 5 + 4n
Step 1: P(1) :  b1 = 5 + 4 = 9 ⇒ 9 = 9 which is true.
Step 2: Assume P(k) is true for some k∈N ⇒ P(k): bk = 5 + 4k is true..
Step 3: Given that:
P(k) = 4 + bk – 1
⇒ P(k + 1) = 4 + bk + 1 – 1
⇒ P(k + 1) = 4 + bk = 4 + 5 + 4k
⇒ P(k + 1) = 5 + 4(k + 1) which is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Q.19. A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk =
for all natural numbers, k ≥ 2. Show that dn =for all n ∈ N.
Ans.
Given that: d1 = 2 and dk =
Let P(n) :
Step 1: P(1) :which is true for P(1).
Step 2: P(k) :Let it be true for P(k).
Step 3: Given that: dk =

Which is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Q.20. Prove that for all  n ∈ N
cos α + cos (α + β) + cos (α + 2β) + ... + cos (α + (n – 1) β)

Ans.
Let P(n) : cos a + cos (a + b) + cos (a + 2b) + ... + cos [a + (n – 1)b]

Step 1: P(1) :cos α =
which is true for P(1)
Step 2: P(k) : cos α + cos (α +β) + cos (α + 2β) + ... + cos [α + (k – 1)β]
Let it be true.
Step 3: P(k + 1) : cos α + cos (α + b) + cos (α + 2b) + ... + cos [α + (k – 1)b]
+ cos [ α + (k + 1 – 1)b]

(from Step 2)

[∴ 2 cos A sin B = sin (A + B) – sin (A – B)]

which is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Q.21. Prove that, cos θ cos 2θ cos22θ ... cos2n–1θfor all n ∈ N.
Ans.
Let P(n) : cos θ.cos 2θ.cos 22θ…cos 2n – 1θ    ,∀n ∈ N.
Step 1: P(1) :
⇒ cos θ = cos θ which is true for P(1)
Step 2: P(k) : cos θ.cos 2θ.cos 22θ ... cos 2k – 1 θ =
Let it be true for P(k).
Step 3: P(k + 1) : cos θ.cos 2θ.cos 22θ ... cos 2k – 1θ. cos 2(k + 1)  1θ

[∵ 2 sin θ cos θ = sin 2θ]
which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

Q.22. Prove that, sin θ + sin 2θ + sin 3θ + ... + sin nθ
for all n ∈ N.
Ans.
Let p(n) : sin θ + sin  + sin  +  + sin

Step 1: P(1) : sin θ =
∴ sin θ = sin θ which is true for P(1).
Step 2: P(k) : sin θ + sin 2θ + sin 3θ + ... + sin kθ
. Let it be true for P(k).
Step 3: P(k + 1) : sin θ + sin 2θ + sin 3θ + ... + sin (k + 1)θ

which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

Q.23. Show thatis a natural number for all n ∈ N.
Ans.
Let P(n)  :
Step 1: P(1)  :
Which is true for P(1).
Step 2: P(k) . Let it be true for P(k) and let

Step 3: P(k + 1) =

= λ + k4 + 2k3 + 3k2 + 2k + 1     [from Step 2]
= positive integers = natural number
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

Q.24. Prove that, for all natural numbers  n > 1.
Ans.
Let P(n) :
Step 1: P(2) :
which is true for P(2).
Step 2: P(k) :.Let it be true for P(k).
Step 3: P(k + 1) :
Since

Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

Q.25. Prove that number of subsets of a set containing n distinct elements is
2n, for all n ∈ N.
Ans.
Let P(n) : Number of subsets of a set containing n distinct
elements is 2n, ∀ n ∈ N
Step 1: It is clear that P(1) is true for n = 1. Number of subsets
= 21 = 2. Which is true.
Step 2: P(k) is assumed to be true for n = k. Since the number of subsets = 2k.
Step 3: P(k + 1) =  2k + 1
We know that if one number (i.e., element) is added to the elements of a given set,
the number of subsets become double.
∴ Number of subsets of set having (k + 1) distinct elements
= 2 x 2k = 2k +  1 which is true for P(k + 1).
Hence P(k + 1) is true
whenever P(k) is true.

Objective Type Questions
Q.26. If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N,
then the least positive integral value of k is
(A) 5
(B) 3
(C) 7
(D) 1
Ans.
Let P(n) = 10n + 3.4n + 2 + k is divisible by 9, ∀ n ∈ N
P(1) = 101 +  3.41 + 2 + k = 10 + 3.64 + k
= 10 + 192 + k = 202 + k must be divisible by 9.
If (202 + k) is divisible by 9 then k must be equal to 5
202 + 5 = 207 which is divisible by 9

So, the least positive integral value of k = 5.
Hence, the correct option is (a).

Q.27. For all n ∈ N, 3.52n+1 + 23n+1 is divisible by
(A) 19
(B) 17
(C) 23
(D) 25
Ans.
Let P(n) : 3.52n + 1 + 23n + 1
For P(1) : 3.52.1 + 1 + 23.1 + = 3.53 + 24 = 3(125) + 16 = 375 + 16
= 391 = 23 x 17
So it is divisible by 17 and 23 both.
Hence, the correct option is (b) and (c).

Q.28. If xn – 1 is divisible by x – k, then the least positive integral value of k is (A) 1
(B) 2
(C) 3
(D) 4
Ans.
Let P(n) = xn – 1 is divisible by x – k.
P(1) = x – 1 is divisible by x – k.
Since k = 1 is the possible least integral value of k.
Hence, the correct option is (a).

Fill in the Blanks
Q.29. If P(n) : 2n < n!, n ∈ N, then P(n) is true for all n ≥ ________.
Ans.
Given that P(n) : 2n < n!,  n ∈ N
For n = 1 , 2 < 1 (Not true)
For n = 2 , 2 × 2 < 2!  4 < 2  (Not true) For n  = 3 , 2 × 3 < 3!  6 < 3.2.1  6 < 6 (Not true) For  n = 4 , 2 × 4 < 4!  8 < 4.3.2.1  8 < 24 (true)
For n = 5 , 2 × 5 < 5!  10 < 5.4.3.2.1 ⇒10 < 120(true)
So, P(n) is true for n  4.
Hence, the value of the filler  is 4.

State whether the following statement is true or false. Justify.
Q.30. Let P(n) be a statement and let P(k) ⇒ P(k + 1), for some natural number k, then P(n) is true for all n ∈ N.
Ans.
Given that: P(k) ⇒ P(k + 1)
P(1)⇒ P(2) which is not true.
Hence, the statement is ‘False’.

The document NCERT Exemplar: Principle of Mathematical Induction | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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## Mathematics (Maths) Class 11

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## FAQs on NCERT Exemplar: Principle of Mathematical Induction - Mathematics (Maths) Class 11 - Commerce

 1. What is the principle of mathematical induction?
Ans. The principle of mathematical induction is a method used to prove statements involving natural numbers. It consists of two steps: the base step, where the statement is proven for a specific value of the natural number, and the induction step, where it is proven that if the statement holds for one natural number, it also holds for the next natural number.
 2. How is the principle of mathematical induction applied in solving mathematical problems?
Ans. The principle of mathematical induction is applied by first proving the statement for the base case, usually the smallest value of the natural number. Then, assuming that the statement holds for a certain natural number, it is proven that it also holds for the next natural number. By repeating this process, the statement is proven to be true for all natural numbers.
 3. What are some common mistakes to avoid when using the principle of mathematical induction?
Ans. Some common mistakes to avoid when using the principle of mathematical induction include: - Forgetting to prove the base case: It is essential to prove the statement for the smallest value of the natural number. - Assuming the statement is true for all natural numbers without proving the induction step: Each step must be proven individually. - Skipping intermediate steps: It is necessary to show the transition from one natural number to the next in the induction step.
 4. Can the principle of mathematical induction be used to prove statements involving real numbers?
Ans. No, the principle of mathematical induction is specifically designed for proving statements involving natural numbers. It relies on the concept of a successor, which is not applicable to real numbers. Different methods, such as proof by contradiction or direct proof, are used to prove statements involving real numbers.
 5. Are there any limitations to the principle of mathematical induction?
Ans. Yes, there are limitations to the principle of mathematical induction. It can only be used to prove statements that hold for all natural numbers. It cannot be used to prove statements that hold for a subset of natural numbers or for infinite sets. Additionally, it cannot be used to prove statements involving real numbers or other mathematical objects beyond the scope of natural numbers.

## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests

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