## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

**Short Answer** **Type Questions **

**(Score 3)**

Question 1.

a P (n): 2^{3n} -1 is an integral multiple of 7,then find P(1)?

b. P (n): 7^{n} – 3^{n} is divisible by 4, then find P(2).

Answer:

a. Let P(n): 2^{3n} -1 is an integral multiple of 7.

For n= 1, P(n) = 2^{3×1} -1

= 2^{3} -1 = 8 -1 =7

b. Let P(n) = 7^{n} – 3^{n} is divisible by 4

For n = 2, P(2) = 7^{2} – 3^{2} = 49 – 9 = 40

Question 2.

a P (n): 5^{n} -1 is an integral multiple of 5, then P(5) =…………..

b. P (n): 5^{2n+2} – 24n – 25 is divisible by 576, then P (0) =

Answer:

a. Let P(n) = 5^{n} -1 is an integral multiple of 5 form= 5, P(5) = 5^{5} – 1 =3125 -1 =3124

b. Let P(n) = 5^{2n+2} – 24n – 25 is divisible by 576

For n = 0,P(0) = 5^{2}-25 = 25-25 = 0

Question 3.

a P (n): n (n+1) (n+2) is divisible by 6, then find P (3).

b Prove the following by using the principle of mathematical induction for all n∈N: n^{2} + n is even.

Answer:

a. Let P(n) = n (n+1) (n+2) is an integral multiple of 5

for = 3, P(3) = 3(3+ 1) (3 +2)

= 3×4×5 = 60

b. P (1): 1^{2}+ 1 is even

P (k): k^{2} + k is even be given Now(k+ l)^{2} + (k+ 1)

= k^{2} + 2k+ 1 +k+ 1

= (k^{2} + k) + (k + 1) which is even

∴ P(k+ 1) is true

Question 4.

a Prove the following by using the principle of mathematical induction for all n ∈N: 2^{n}<3^{n }b. 1^{3} + 2^{3} + 3^{3} +…………… + n^{3} =………………….

Answer:

a. P(1) is 2 < 3, which is tme. Assume P(r) is true, hence find out P(r + 1)

P(r+1) : 2^{r+1} <3^{r+1}

Now 2^{r+1} ^{ =} 2^{r}.2<3^{r}.2 (∴2^{r}<3^{r})

i.e.,2^{r+}^{1}<3^{r}.2<3^{r}. 3, ∴ 2^{r+1}<3^{r+1 }Question 5.

Prove that 1^{3} + 2^{3} + 3^{3} +……………n^{2 }=………………….

Answer:

Question 6.

Consider the statement P(n): 2^{3n} -1 is divisible by 7, ∀n∈N

i. Is the statement P(1) true? Justify your answer.

ii. If P(k) is true,show that P(k + 1) is also true.

iii. Do you conclude that the given statement is true?

Answer:

i. Yes. P(1):2^{3}-l = 7is divisible by 7.

ii. 2^{3(k+1)} = 2^{3k}.2^{3} – 1 = 8.2^{3k}-1 = 8.2^{3k}-8 + 7 = 8 (2^{3k} -1) + 7 which is divisible by 7, since P (k) is true as given. Hence P (k + 1) is also true.

iii. The given statement is also true, by applying the principle of mathematical induction.

Question 7.

Prove the following by using the principle of mathematical induction for all n∈N

Answer:

Question 8.

Prove the following by using the principle of mathematical induction for all

n ∈ N: 1^{2} + 2^{2} + 5^{2} +……………+ (2k-1)^{2} + [2(k+1)-1]^{2 }

Answer:

Question 9.

Prove the following by using the principle of mathematical induction for all n ∈ N:

a. n (n +1) (n + 2) is divisible by 6

b. n^{3} + (n +1)^{3} + (n + 2)^{3} is divisible by 9

Answer:

a. Proof of P(r + 1):(r+ 1)(r + 2)(r + 3)

= r(r+ 1)(r+2) + 3 (r+ 1)(r+2)

= r(r+ 1)(r + 2) + 3.2 M

(r+ 1) (r + 2) is even = r(r+1)(r + 2) + 6M

which is divisible by 6, since r (r + 1) (r+ 2) is divisible by 6.

b. Proof of P(k+ 1):(k+ 1)^{3} + (k+2)^{3} + (k+3)^{3 }= (k+ 1)^{3} + (k+2)^{3} + k^{3} + 9k^{2} + 27k + 27

= [k^{3} + (k + l)^{3} + (k+2)^{3}] + 9 [k^{2} + 3k + 3]

The first term is divisible by 9 (from P (k))

∴ P(k+ 1) is true.

Question 10.

Prove 5^{2n + 2} – 24n – 25 is divisible by 576 by using the principle of mathematical induction for all n ∈ N:

Answer:

P (k): 5^{2k+2} – 24k – 25 is divisible by 576

Now 5^{2k-4}– 24k -49

= 25.5^{2k+2} – 24k – 49

= 25 [5^{2k+2} – 24k – 25] + 576k + 576

= 25 (5^{k+2} – 24k – 25) + 576 (k + 1)

∴ P(k+ 1) is true

Question 11.

a. Prove the following by using the principle of mathematical induction for all n ∈ N:

2^{1} + 2^{2} +…. + 2^{n}= 2^{n+1}– 2

b. Prove the following by using the principle of mathematical induction for all

n ∈ N: 1.2 +3.4+5.6+… +(2n-1)(2n)

Answer:

Question 12.

Prove 1.2 + 2.2^{2} + 2.3^{2} +……………. + n.2^{n} = (n-1) 2^{n+1} + 2 by using the principle of mathematical induction for all n∈N.

Answer:

Let P(n): 1.2 + 2.2^{2} + 3.2^{3} + ….+n.2^{n}

= (n -1 )2^{n+1} + 2

i. P(1): 1.2 =2 = (1 -1 )2 1+1 +2=2, which is true.

Thus P(1) is true.

ii. Assume that P(m) is true

i.e.., P(m)= 1.2 + 2.2^{2} + 3.2^{3} + + m.2^{m }= (m-1 )2^{m+}^{1}+2.

iii. LHS of P(m+1)

= 1.2 + 2.2^{2} + 3.2^{3} +…. + m.2^{m} + (m+1)2^{m+1 }= (m-1)2^{m+1}+2+ (m+1)2^{m+1} (Using (ii))

= 2^{m+1} [m-1 +m+1] +2 = 2^{m+1} (2m) + 2 = m. 2^{m+2}+ 2

= RHS of P(m+1)

∴ P(m+ 1) is true.

i.e.,P(m)is true ⇒ P(m+1)is true.

∴ By the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 13.

Using the Principle of Mathematical Induction, prove that 10^{2n+1} is divisible by 11.

Answer:

**Step I:** Let P(n) be the given statement. Then,

P(n): 10^{2n+1}+ 1 is divisible by 11.

**Step II:** For n=1, we have

p(1): 10^{2–}^{1} + 1 = 10 + 1 = 11

which is divisible by 11.

Thus, P(1) is true.

**Step III:** For n = k, assume that P(k) is true.

i.e., P(k):10^{2k- l} +1 is divisible by 11.

10^{2k-1} + 1 = 11

where,m ∈ N

**Step IV.** For n = k+1, we have to prove that P(k+1) is true, whenever P(k) is true,

i.e., P(k+1):102(k+1) -1 + 1 is divisible by 11

Now, 10^{2(k+1)-1} + 1 = 10^{2k+2-1} + 1

= 10^{2k+1} +1 = 10^{2k+2-1}+ 1

= 100(10^{2k-1 }+ 1))+100-99

= 100(10^{2k-1} + 1)-99

= 11 (100m -9), which is divisible by 11.

Thus, P(k+1) is divisible by 11 whenever P(k) is divisible by 11.

Hence, by principle of Mathematical Induction, P(n) is true for all n∈N.

**Long Answer Type Questions**

**(Score – 6)**

Question 1.

If P (n) is the statement ‘4n < 2^{n}’

i. Verify that P(1), P(2), P(3), P(4) and P(5) are true.

ii. Can you say that P(n) will be true for all natural numbers? Give reason.

iii. If P (k) is true, show that P (k +1) is true fork >5.

iv. Prove that P (n): 4n < 2^{n}, ∀ n > 5

Answer:

i. P(1):4(1)<2^{1}, P (2): 4 (2)<2^{2},

P (3): 4 (3) < 2^{3}, P (4): 4 (4) < 2^{4}, are not true

P (5): 4 (5) < 2^{5} is true.

ii. From the above discussion it is evident that P (n) is not true for all natural numbers.

iii. Given P(k) is true.

∴ P(k):4k<2^{k},k>5.

Now 2^{k–}^{1} = 2^{k}. 2 > 4k. 2

= 4(2k) = 4(k + k)>4(k+ 1)

⇒ 4 (k+ 1) < 2^{k+1 }∴ P (k+ 1) is true.

Question 2.

Using (i) and (iii) and applying principle of mathematical induction, P (n) is true for all n>5. Using the principle of Mathematical Induction, prove that 1.3 + 2.3^{2} + 3.3^{3} +……………. + n. 3”

Answer:

Therefore, P(k+1) is true, whenever P(k) is true.Hence, by principle of Mathematical Induction, P(n) is true for all n ∈N.

Question 3.

a. Prove by the Principle of Mathematical Induction that logx^{n} = nlogx.

b Prove by the Principle of Mathematical Induction that 3^{n} > 2^{n} .

Answer:

**a. Step I:**

Let P(n) be the statement given by

P(n): logx^{n} = nlogx

**Step II:**

For n=l, we have

LHS = logx^{1 }=1.logx RHS

= logx Thus, P(1) is true.

**Step III:** For n = k assume that P(k) is true,

i.e., P(k): logx^{k}=klogx

**Step IV:** For n = k+1, we have to show that

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

i.e., P(k+1): log x^{k+1} = (k+1) logx

LHS = logx^{k+1 }= log(x^{k}.x) = logx^{k}+logx

= klogx + logx = (k+1 )logx = RHS

Thus, P(k+1) is true, whenever P(k) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n e N.

b. **Step I:** Let P(n) be the statement given by

P(n):3n>2

**Step II:** Forn= 1, we have 3^{1}>2^{1 }3 >2, which is true, Therefore, P(1) is true.

**Step III:** For n = k+1, we have to show that P(k+1) is true, whenever P(k) is true.

i.e.,P(k) :3k>2k

**Step IV:** For n= k+1, we have to show that

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

i.e.,P(k+1):3^{k+1}>2^{k+1 }Now, 3^{k+1} = 3^{k}.3 > 2^{k}.3

But 2^{k}.3> 2^{k}.2

2^{k}.3 > 2^{k+1 }⇒ 3^{k}.3>2^{k+1 }3^{k+1}>2^{k+1 }Thus, P(k+1) is true, whenever P(k) is true.

Hence, by Principle of Mathematical Induction, P(n) is true for all n∈N.

Question 4.

Prove by the Principle of Mathematical Induction that x^{2n}– y ^{2n} is divisible by x + y> for all n ∈ N

Answer:

**Step I:** Let the given statement be P(n). Then,

P(n): x^{2n} – y^{2n} is divisible by (x+y).

**Step II:** For n = 1, we have

P( 1) = x^{2n} – y^{2}” = x^{2} – y^{2} = (x+y)(x – y) which is divisible by (x + y).

Thus, P(n) is true for n = 1.

**Step III:** For n = k, assume that P(k) is true, i.e., P(k) = x^{2k} – y^{2k} is divisible by x+y.

Then, x^{2k} – y^{2k}=m(x+y) x^{2k} = m^{<x+y)}+y^{2k}

**Step IV:** For n = k+1, we have to show that P(k+1) is true whenever P(k) is true.

i.e., P(k+1) = x^{2(k+1)} – y^{2(k+1>} is divisible by (x+y).

Now, P(k+1) = x^{2(k+1)} – y^{2(k+1) }= x^{2(k+1)} – y^{2(k+1) }= x^{2} .x^{2k}– y^{2}. y^{2k }= x^{2} [ m(x+y) + y^{2k})] – y^{2}. y^{2k }= m(x+y)x^{2} + x^{2}. y^{2k} – y^{2}. y^{2k }= m(x+y)x^{2} +(x+y)(x -y)y^{2k }= (x+y)[mx^{2} + (x-y)y^{2k}]

which is divisible by (x+y).

Thus, x^{2(k+1)} – y^{2(k+1)} is divisible by (x+y), i.e., P(k+1) is true, whenever P(k) is true.

Hence, by Principle of Mathematical Induction, P(n) is true for all n ∈N.

Question 5.

Prove by the Principle of Mathematical Induction that

Answer:

Therefore, P(k+1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 6.

Consider the statement

P(n) : 1×2 + 2×3+3×4 + ….+ n(n+1)

a. Verify that P(3) is true.

b. Prove that statement P(n) is true for all natural numbers using mathematical induction.

Answer:

Therefore, P(k+1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n∈N.

**NCERT Question and Answers**

Question 1.

Consider the statement P(n): 1^{2} + 2^{2} + 3^{2} +………..

∀n ∈N.

1. Is P(1) true? Explain.

2. Find P (k) and P (k+1).

3. If P (k) is true, prove that P (k+1) is true.

Answer:

.

P (n) is true. From the above discussions, applying the principle of mathematics induction, we get P (n) is true ∀n∈N.

Question 2.

(1 +x)^{n}>1 +nx for n>1,x>-1

arts:

P (2): (1 + x)^{2} > 1 + 2x.

Now(1 +x)^{2}= 1 +2x + x^{2}> 1 +2x

∴ P (2) is true. Assume P (k) is true

i.e., P (k) : (1 +x)^{k}> 1 +kx

Proof of P (k+1): (1 +x)^{k+I} = (1 +x)^{k}(l +x)> (1 + k x) (1 + x), from P (k)

= 1 + x + kx + kx^{2} = 1 + (k + l)x + kx^{2},

kx^{2}>0> 1 +(k+ 1)x, P(k+ 1) is true

Question 3.

7^{n} + 3.5^{n} – 5 is divisible by 24

Answer:

2.7^{k} + 3,5^{k+1} – 5 is divisible by 24

Proof of P (k+1) 2.7^{k+1} + 3.5^{k+1} – 5

= 2.7^{k}(6+ l) + 3.5^{k}(4+ 1)-5

= 12(7^{k}+5^{k}) + [2.7^{k} + 3.5^{k}-5]

7^{k} and 5^{k} are odd numbers, therefore (7^{k} + 5^{k}) will be even and 12 (7^{k} + 5^{k}) will be divisible by 24, the second term will also be divisible by 24, from P (k).

P(k+ 1) is true Hence P(n) is true, ∀n∈N.

Question 4.

1^{2} + 2^{2}+ ….+ n^{2}

Answer:

Question 5.

Prove the following by using the principle of mathematical induction for all n ∈ N

Answer:

Question 6.

Prove the following by using the principle of mathematical induction for all n∈N

1.2 + 2.2^{2} + 3.2^{3} +…… + n.2^{n} = (n -1)2^{n} ^{+1} +2

Answer:

P(1) = 1^{3}= 1, which is true

Let P (n): 1.2 + 2.2^{2} + 3,2^{3} +.. + n.2^{n} = (n -1) 2^{n+1 }+2

P(1): 1.2 = 0 + 2, which is true P(1) is true.

Assume P (k) is true.

P (k) = 1.2 + 2.2^{2} + 3.2^{3} +….. + k.2^{k} = (k -1 )2^{k+1 }+2———– (1)

Now 1.2 + 2.2^{2} + 3,2^{3} +… + k.2^{k} + (k +1 )2^{k+1} +2

= (k – 1)2^{k+1} + 2 + (k + l)2^{k+1},

by (1) =2^{k+1}[k-1+k+1] + 2 = 2k2^{k}-‘ + 2

= k.2^{k+2} + 2 P(k+1)is true.

Hence P (n) is true,∀n∈N

Question 7.

Prove the following by using the principle of mathematical induction for all n∈N

Answer:

Question 8.

Prove the following by using the principle of mathematical induction for all n∈N

Answer:

Question 9.

Prove the following by using the principle of mathematical induction for all n∈N n (n +1) (n + 5) is a multiple of 3

Answer:

P (1) is true. Assume P (k) is true. k(k+ 1) (k+ 5) = 3 M

To prove P (k + 1) is true (k- 1) (k + 2) (k + 6) = (k -1) (k^{2} +8 k+12)

= (k+ 1)(k^{2} + 5k + 3k+ 12)

= (k+ 1) [k (k + 5) + 3 (k + 4)]

= k(k- 1)(k-5) + 3(k- 1)(k + 4)

= 3 M-3 (k= 1)(k + 4)

= 3[M-(k-1)(k + 4)]

∴ P (k- 1) is true.

Question 10

10^{2n -1} + 1 is divisible by 11

Answer:

Question 11.

3^{2n+2} – 8n – 9 is divisible by 8

Answer: