NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction

In this chapter, we provide NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction pdf, free NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 11
SubjectMaths
ChapterChapter 4
Chapter NamePrinciple of Mathematical Induction
CategoryNCERT Exemplar

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction

Short Answer Type Questions
Q1. Give an example of a statement P(n) which is true for all n≥ 4 but P(l), P(2) and P(3) are not true. Justify your answer.

Sol. Consider the statement P(n): 3n < n!

For n = 1, 3 x 1 < 1!, which is not true
For n = 2, 3 x 2 < 2!, which is not true
For n = 3, 3 x 3 < 3!, which is not true
For n = 4, 3 x 4 < 4!, which is true
For n = 5, 3 x 5 < 5!, which is true

Q2. Give an example of a statement P(n) which is true for all Justify your answer.

Instruction for Exercises 3-16: Prove each of the statements in these Exercises by the Principle of Mathematical Induction.

Q3. 4n – 1 is divisible by 3, for each natural number
Sol: Let P(n): 4n – 1 is divisible by 3 for each natural number n.
Now, P(l): 41 – 1 = 3, which is divisible by 3 Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 4k – 1 is divisible by 3
or               4k – 1 = 3m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 4k+1 – 1
= 4k-4-l
= 4(3m + 1) – 1  [Using (i)]
= 12 m + 3
= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q4. 23n – 1 is divisible by 7, for all natural numbers
Sol: Let P(n): 23n – 1 is divisible by 7
Now, P( 1): 23 — 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 23k – 1 is divisible by 7.
or               23k -1 = 7m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 23(k+1)– 1
= 23k.23– 1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q5. n3 – 7n + 3 is divisible by 3, for all natural numbers
Sol: Let P(n): n3 – 7n + 3 is divisible by 3, for all natural numbers n.
Now P(l): (l)3 – 7(1) + 3 = -3, which is divisible by 3.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) = K3 – 7k + 3 is divisible by 3
or K3 – 7k + 3 = 3m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1 ):(k + l)3 – 7(k + 1) + 3
= k3 + 1 + 3k(k + 1) – 7k— 7 + 3 = k3 -7k + 3 + 3k(k + l)-6
= 3m + 3[k(k+l)-2]  [Using (i)]
= 3[m + (k(k + 1) – 2)], which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q6. 32n – 1 is divisible by 8, for all natural numbers
Sol: Let P(n): 32n – 1 is divisible by 8, for all natural numbers n.
Now, P(l): 32 – 1 = 8, which is divisible by 8.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): 32k – 1 is divisible by 8
or               32k -1 = 8m, m ∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 32(k+1)– l
= 32k • 32 — 1
= 9(8m + 1) – 1     (using (i))
= 72m + 9 – 1
= 72m + 8
= 8(9m +1), which is divisible by 8 Thus P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q7. For any natural number n, 7n – 2n is divisible by 5.
Sol: Let P(n): 7n – 2n is divisible by 5, for any natural number n.
Now, P(l) = 71-21 = 5, which is divisible by 5.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.

.’.  P(k) = 7k -2k is divisible by 5
or  7k – 2k = 5m, m∈ N                                                                           (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 7k+1 -2k+1
= 7k-7-2k-2
= (5 + 2)7k -2k-2
= 5.7k + 2.7k-2-2k
= 5.7k + 2(7k – 2k)
= 5 • 7k + 2(5 m)     (using (i))
= 5(7k + 2m), which divisible by 5.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q8. For any natural number n, xn -yn is divisible by x -y, where x and y are any integers with x ≠y
Sol:
Let P(n) : xn – yn is divisible by x – y, where x and y are any integers with x≠y.
Now, P(l): x1 -y1 = x-y, which is divisible by (x-y)
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): xk -yk is divisible by (x – y)
or   xk-yk = m(x-y),m ∈ N …(i)
Now, we have to prove that P(k + 1) is true.
P(k+l):xk+l-yk+l
= xk-x-xk-y + xk-y-yky
= xk(x-y) +y(xk-yk)
= xk(x – y) + ym(x – y)  (using (i))
= (x -y) [xk+ym], which is divisible by (x-y)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q9. n3 -n is divisible by 6, for each natural number n≥
Sol: Let P(n): n3 – n is divisible by 6, for each natural number n> 2.
Now, P(2): (2)3 -2 = 6, which is divisible by 6.
Hence, P(2) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): k3 – k is divisible by 6
or    k3 -k= 6m, m∈ N       (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): (k+ l)3-(k+ 1)
= k3+ 1 +3k(k+ l)-(k+ 1)
= k3+ 1 +3k2 + 3k-k- 1 = (k3-k) + 3k(k+ 1)
= 6m + 3 k(k +1)  (using (i))
Above is divisible by 6.   (∴ k(k + 1) is even)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 2.

Q10. n(n2 + 5) is divisible by 6, for each natural number
Sol: Let P(n): n(n2 + 5) is divisible by 6, for each natural number.
Now P(l): 1 (l2 + 5) = 6, which is divisible by 6.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): k( k2 + 5) is divisible by 6.
or K (k2+ 5) = 6m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(K+l):(K+l)[(K+l)2 + 5]
= (K + l)[K2 + 2K+6]
= K3 + 3 K2 + 8K + 6
= (K2 + 5K) + 3 K2 + 3K + 6 =K(K2 + 5) + 3(K2 + K + 2)
= (6m) + 3(K2 + K + 2)        (using (i))
Now, K2 + K + 2 is always even if A is odd or even.
So, 3(K2 + K + 2) is divisible by 6 and hence, (6m) + 3(K2 + K + 2) is divisible by 6.
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q11. n2 < 2n, for all natural numbers n ≥
Sol: Let P(n): n2 < 2n for all natural numbers n≥ 5.
Now P(5): 52 < 25 or 25 < 32, which is true.
Hence, P(5) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): k2 < 2k  (i)
Now, to prove that P(k + 1) is true, we have to show that P(k+ 1): (k+ l)2 <2k+1
Using (i), we get
(k + l)2 = k2 + 2k + 1 < 2k + 2k + 1         (ii)
Now let, 2k + 2k + 1 < 2k+1     (iii)
∴ 2k + 2k + 1 < 2 • 2k
2k + 1 < 2k, which is true for all k > 5 Using (ii) and (iii), we get (k + l)2 < 2k+1 Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 5.

Q12. 2n<(n + 2)! for all natural numbers
Sol: Let P(n): 2n < (n + 2)! for all natural numbers n.
P( 1): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.
Hence,P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) :2k<(k + 2)!  (i)
To prove that P(k + 1) is true, we have to show that
P(k + 1): 2(k+ 1) < (k + 1 + 2)!
or 2(k+ 1) < (k + 3)!
Using (i), we get
2(k + 1) = 2k + 2<(k+2) !  +2  (ii)
Now let, (k + 2)! + 2 < (k + 3)!  (iii)
=>  2 < (k+ 3)! – (k+2) !
=> 2 < (k + 2) ! [k+ 3-1]
=>2<(k+ 2) ! (k + 2), which is true for any natural number.
Using (ii) and (iii), we get 2(k + 1) < (k + 3)!
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

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Q14. 2 + 4 + 6+… + 2n = n2 + n, for all natural numbers
Sol: Let P(n) :2 + 4 + 6+ …+2 n = n2 + n
P(l): 2 = l2 + 1 = 2, which is true
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + .,.+2k = k2 + k  (i)
Now, we have to prove that P(k + 1) is true.
P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)
= k2 + k + 2(k+ 1)  [Using (i)]
= k2 + k + 2k + 2
= k2 + 2k+1+k+1
= (k + 1)2 + k+ 1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q15. 1 + 2 + 22 + … + 2n = 2n +1 – 1 for all natural numbers
Sol: Let P(n): 1 + 2 + 22 + … + 2n = 2n +1 – 1, for all natural numbers n
P(1): 1 =20 + 1 — 1 = 2 — 1 = 1, which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.

P(k): l+2 + 22+…+2k = 2k+1-l              (i)

Now, we have to prove that P(k + 1) is true.

P(k+1): 1+2 + 22+ …+2k + 2k+1
= 2k +1 – 1 + 2k+1  [Using (i)]
= 2.2k+l– 1 = 1
= 2(k+1)+1-1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q16. 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers
Sol: Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 x 1 – 1) = 1, which is true.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k):l+5 + 9 +…+(4k-3) = k(2k-1)  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 1 + 5 + 9 + … +  (4k- 3) + [4(k+ 1) – 3]
= 2k2 -k+4k+ 4-3
= 2k2 + 3k + 1
= (k+ 1)( 2k + 1)

= (k+l)[2(k+l)-l]

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

So, by the principle of mathematical induction P(n) is true for any natural number n.

Long Answer Type Questions
Q17. A sequence ax, a2, a3, … is defined by letting a1=3 and ak = 7ak1 for all natural numbers k≥ Show that an = 3 • 7 n-1 for all natural numbers.
Sol: We have a sequence ax, a2, a3… defined by letting a, = 3 and ak = 7ak1, for all natural numbers k≥2.
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Q18. A sequence b0, b1, b2, … is defined by letting b0 = 5 and bk = 4 + bk1, for all natural numbers Show that bn = 5 + 4n, for all natural number n using mathematical induction.
Sol. We have a sequence b0, b1, b2,… defined by letting b0 = 5 and bk = 4 + bk1,, for all natural numbers k.

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So, by the principle of mathematical induction P(n) is true for any natural number rt,n> 1.

Q25. Prove that number of subsets of a set containing n distinct elements is 2″, for all n ∈
Sol: Let P(n): Number of subset of a set containing n distinct elements is 2″, for all ne N.
For n = 1, consider set A = {1}. So, set of subsets is {{1}, ∅}, which contains 21 elements.
So, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
P(k): Number of subsets of a set containing k distinct elements is 2k To prove that P(k + 1) is true,
we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2k+1
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing (k+ 1) distinct elements = 2×2k = 2k+1
So, P(k + 1) is true. Hence, P(n) is true.

All Chapter NCERT Exemplar Problems Solutions For Class 11 Maths

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