**NCERT Exemplar Problems Class 11 Mathematics 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. 4 ^{n} – 1 is divisible by 3, for each natural number**

**Sol:**Let P(n): 4

^{n}– 1 is divisible by 3 for each natural number n.

Now, P(l): 4

^{1}– 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): 4

^{k}– 1 is divisible by 3

or 4

^{k}– 1 = 3m, m∈ N (i)

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

P(k+ 1): 4

^{k+1}– 1

= 4

^{k}-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. 2 ^{3n} – 1 is divisible by 7, for all natural numbers**

**Sol:**Let P(n): 2

^{3n}– 1 is divisible by 7

Now, P( 1): 2

^{3}— 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): 2

^{3k}– 1 is divisible by 7.

or 2

^{3k}-1 = 7m, m∈ N (i)

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

P(k+ 1): 2

^{3(k+1)}– 1

= 2

^{3k}.2

^{3}– 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. n ^{3} – 7n + 3 is divisible by 3, for all natural numbers**

**Sol:**Let P(n): n

^{3}– 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) = K

^{3}– 7k + 3 is divisible by 3

or K

^{3}– 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

= k

^{3}+ 1 + 3k(k + 1) – 7k— 7 + 3 = k

^{3}-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. 3 ^{2}^{n} – 1 is divisible by 8, for all natural numbers**

**Sol:**Let P(n): 3

^{2}

^{n}– 1 is divisible by 8, for all natural numbers n.

Now, P(l): 3

^{2}– 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): 3

^{2k}– 1 is divisible by 8

or 3

^{2k}-1 = 8m, m ∈ N (i)

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

P(k+ 1): 3

^{2(k+1)}– l

= 3

^{2k}• 3

^{2}— 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, 7 ^{n }– 2^{n} is divisible by 5.**

**Sol:**Let P(n): 7

^{n}– 2

^{n}is divisible by 5, for any natural number n.

Now, P(l) = 7

^{1}-2

^{1}= 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) = 7^{k} -2^{k} is divisible by 5

or 7^{k} – 2^{k} = 5m, m∈ N (i)

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

P(k+ 1): 7^{k+1} -2^{k+1 }= 7^{k}-7-2^{k}-2

= (5 + 2)7^{k} -2^{k}-2

= 5.7^{k} + 2.7^{k}-2-2^{k }= 5.7^{k} + 2(7^{k} – 2^{k})

= 5 • 7^{k} + 2(5 m) (using (i))

= 5(7^{k} + 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, x ^{n} -y^{n} is divisible by x -y, where x and y are any integers with x ≠y** Let P(n) : x

Sol:

^{n }– y

^{n}is divisible by x – y, where x and y are any integers with x≠y.

Now, P(l): x

^{1}-y

^{1}= 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): x

^{k}-y

^{k}is divisible by (x – y)

or x

^{k}-y

^{k}= m(x-y),m ∈ N …(i)

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

P(k+l):x

^{k+l}-y

^{k+l }= x

^{k}-x-x

^{k}-y + x

^{k}-y-y

^{k}y

= x

^{k}(x-y) +y(x

^{k}-y

^{k})

= x

^{k}(x – y) + ym(x – y) (using (i))

= (x -y) [x

^{k}+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. n ^{3} -n is divisible by 6, for each natural number n≥**

**Sol:**Let P(n): n

^{3}– 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): k

^{3}– k is divisible by 6

or k

^{3}-k= 6m, m∈ N (i)

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

P(k+ 1): (k+ l)

^{3}-(k+ 1)

= k

^{3}+ 1 +3k(k+ l)-(k+ 1)

= k

^{3}+ 1 +3k

^{2}+ 3k-k- 1 = (k

^{3}-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(n ^{2} + 5) is divisible by 6, for each natural number**

**Sol:**Let P(n): n(n

^{2}+ 5) is divisible by 6, for each natural number.

Now P(l): 1 (l

^{2}+ 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( k

^{2}+ 5) is divisible by 6.

or K (k

^{2}+ 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)[K

^{2}+ 2K+6]

= K

^{3}+ 3 K

^{2}+ 8K + 6

= (K

^{2}+ 5K) + 3 K

^{2}+ 3K + 6 =K(K

^{2}+ 5) + 3(K

^{2}+ K + 2)

= (6m) + 3(K

^{2}+ K + 2) (using (i))

Now, K

^{2}+ K + 2 is always even if A is odd or even.

So, 3(K

^{2}+ K + 2) is divisible by 6 and hence, (6m) + 3(K

^{2}+ 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. n ^{2} < 2^{n}, for all natural numbers n ≥**

**Sol:**Let P(n): n

^{2}< 2

^{n}for all natural numbers n≥ 5.

Now P(5): 5

^{2}< 2

^{5}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): k

^{2}< 2

^{k}(i)

Now, to prove that P(k + 1) is true, we have to show that P(k+ 1): (k+ l)

^{2}<2

^{k+}

^{1 }Using (i), we get

(k + l)

^{2}= k

^{2}+ 2k + 1 < 2

^{k}+ 2k + 1 (ii)

Now let, 2

^{k}+ 2k + 1 < 2

^{k}

^{+1}(iii)

∴ 2

^{k}+ 2k + 1 < 2 • 2

^{k }2k + 1 < 2

^{k}, which is true for all k > 5 Using (ii) and (iii), we get (k + l)

^{2}< 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,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.

**Q14. 2 + 4 + 6+… + 2n = n ^{2} + n, for all natural numbers**

**Sol:**Let P(n) :2 + 4 + 6+ …+2 n = n

^{2}+ n

P(l): 2 = l

^{2}+ 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 = k

^{2}+ k (i)

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

P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)

= k

^{2}+ k + 2(k+ 1) [Using (i)]

= k

^{2}+ k + 2k + 2

= k

^{2}+ 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 + 2 ^{2} + … + 2^{n} = 2^{n} ^{+1} – 1 for all natural numbers**

**Sol:**Let P(n): 1 + 2 + 2

^{2}+ … + 2

^{n}= 2

^{n}

^{+1}– 1, for all natural numbers n

P(1): 1 =2

^{0 + 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 + 2^{2}+…+2^{k} = 2^{k+1}-l (i)

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

P(k+1): 1+2 + 2^{2}+ …+2^{k} + 2^{k+1 }= 2^{k +1} – 1 + 2^{k+1} [Using (i)]

= 2.2^{k+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]

= 2k^{2} -k+4k+ 4-3

= 2k^{2} + 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 a _{x}, a_{2}, a_{3}, … is defined by letting a_{1}=3 and a_{k} = 7a_{k}–_{1} for all natural numbers k≥ Show that a_{n} = 3 • 7 ^{n-1} for all natural numbers.**

**Sol:**We have a sequence a

_{x}, a

_{2}, a

_{3}… defined by letting a, = 3 and a

_{k}= 7a

_{k}–

_{1}, for all natural numbers k≥2.

**Q18. A sequence b _{0}, b_{1}, b_{2}, … is defined by letting b_{0} = 5 and b_{k} = 4 + b_{k}–_{1}, for all natural numbers Show that b_{n} = 5 + 4n, for all natural number n using mathematical induction.**Sol. We have a sequence b

_{0}, b

_{1}, b

_{2},… defined by letting b

_{0}= 5 and b

_{k}= 4 + b

_{k}–

_{1},, for all natural numbers k.

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 2^{1} 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 2^{k }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 2^{k+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 = 2×2^{k} = 2^{k+1 }So, P(k + 1) is true. Hence, P(n) is true.