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

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

### Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers

Question 1.
For all n ≥ 1 , prove that
12 + 22 + 32 +……….+ n2 > $$\frac{n^{3}}{3}$$
Let p(n): 12 + 22 + 32 + n2
Put n = 1 ⇒ p(1) = 1 > $$\frac{1}{3}$$ which is true.
Assuming that true for p(k)
p(k): 12 + 22 + 32 +……….+ k2 > $$\frac{k^{3}}{3}$$
Let p(k + 1): 12 + 22 + 32 +…….+ k2 + (k + 1)2 > $$\frac{k^{3}}{3}$$ + (k + 1)2

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 2.
For all n ≥ 1 , prove that 1 + 2 + 3 +…….+ n < $$\frac{1}{8}$$(2n + 1)2
Let p(n): 1 + 2 + 3 +…….+ n ,
Put n = 1 ⇒ p(1) = 1 < $$\frac{9}{8}$$ which is true.
Assuming that true for p(k)
p(k): 1 + 2 + 3 +…….+ k < $$\frac{1}{8}$$(2k + 1)2
Let p(k +1): 1 + 2 + 3 +……..+ k + (k +1) < $$\frac{1}{8}$$ (2k + 1)2 + (k + 1)

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1, prove that p(n): 23n – 1 is divisible by 7.
p(1): 23(1) – 1 = 8 – 1 = 7 divisible by 7, hence true. Assuming that for p(k)
p(k) : 23k – 1 is divisible by 7.
23k – 1 = 7M
P(k + 1): 23(k + 1) – 1 = 23k + 3 – 1
= 23k23 – 1 = 23k × 8 – 1
= 23k × 8 – 8 + 7 = 8(23k – 1) + 7
= 8(7M) + 7
Hence divisible by 7. Therefore by using the principle of mathematical induction true for all n ∈ N.

Question 4.
For all n ≥ 1, prove that p(n): n3 + (n + 1)3 + (n + 2)3 is divisible by 9.
p(1): 1 + 23 + 33 = 1 + 8 + 27 = 36 divisible by 9, hence true. Assuming that true for p(k)
p(k): k3 + (k + 1)3 + (k + 2)3 is divisible by 9.
⇒ k3 + (k + 1)3 + (k + 2)3 = 9M
p(k +1 ): (k + 1)3 + (k + 2 )3 + (k + 3)3
= (k +1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27
= [(k + 1)3 + (k + 2)3 + k3] + 9[k2 + 3k + 3]
= 9M + 9[k2 + 3k + 3]
Hence p(k + 1)divisible by 9. Therefore by using the principle of mathematical induction true for all n ∈ N.

### Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers

Question 1.
For all n ≥ 1, prove that

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 2.
For all n ≥ 1, prove that

Let

Assuming that true for p(k)

Let p(k + 1):

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1 , prove that 1.2.3 + 2.3.4 +………+ n(n + 1)(n + 2) = $$\frac{n(n+1)(n+2)(n+3)}{4}$$.
Let p(n): 1.2.3 + 2.3.4 +……..+ n(n + 1)(n + 2),
Put n = 1
p(1) = $$\frac{1(1+1)(1+2)(1+3)}{4}$$ = 6 which is true.
Assuming that true for p(k)
p(k): 1.2.3 + 2.3.4 +……..+ k(k + 1)(k + 2) = $$\frac{k(k+1)(k+2)(k+3)}{4}$$,
Let p(k + 1)

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 4.
For all n ≥ 1, prove that

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 5.
For all n ≥ 1 , prove that p(n): n(n + 1 )(n + 5) is divisible by 3.
p(1): 1(1 + 1)(1 + 5) = 12divisible by 3, hence true. Assuming that true for p(k)
p(k): k(k + 1)(k + 5) is divisible by 3.
k(k + 1)(k + 5) = 3M
p(k + 1): (k + 1)(k + 2)(k + 6)
= (k + 1)(k2 + 8k + 12)
= (k + 1)(k2 + 5k + 3k +12)
= (k + 1)[k(k + 5) + 3(k + 6)]
= [k(k + 1)(k + 5) + 3(k + 1)(k + 6)]
= [3M + 3(k + 1)(k + 6)]
= 3[M + (k + 1)(k + 6)]
Hence divisible by 3. Therefore by using the principle of mathematical induction true for all n ∈ N.

Question 6.
For all n ≥ 1 , prove that p(n): 2.7n + 3.5n – 5 is divisible by 24.