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Chapter 12: Mathematical Induction – Exercise 12.2

Mathematical Induction – Exercise 12.2 – Q.1

Mathematical Induction – Exercise 12.2 – Q.1

For n = 1

LHS of P(n) = 1

since, LHS = RHS

⟹ p(n)is true for n = 1

Let P(n)  be true for n = k, so

Now

⟹p(n)  is true for n = k + 1

⟹p(n)  is true for all n ϵ N

so, by the principle of mathmematical induction

 

Mathematical Induction – Exercise 12.2 – Q.2

Mathematical Induction – Exercise 12.2 – Q.2

For n = 1

1 + 1

 ⟹ p(n)  is true for n = 1

Let p(n)  is true for n = k, so

We have to show that p(n)is true for n = k + 1

⟹ p(n) is true for n = k + 1

⟹ p(n) is true for all n ϵ N by PMI

 

Mathematical Induction – Exercise 12.2 – Q.3

Mathematical Induction – Exercise 12.2 – Q.3

For n = 1

1 = 1

⟹p(n) is true for n = 1

Let p(n) is true for n = k

we have to show P(n) is true for n = k + 1

Now,

⟹ p(n) is true for n = k + 1

⟹ p(n) is true for all n ϵ N by PMI

 

Mathematical Induction – Exercise 12.2 – Q.4

Mathematical Induction – Exercise 12.2 – Q.4

For n = 1

⟹ p(n)  is true for n = 1

Let p(n)  is true for n = k, so

we have to show that

Now,

⟹ p(n)  is true for n =k + 1

⟹ p(n)  is true for all n ϵ N by PMI

 

Mathematical Induction – Exercise 12.2 – Q.5

Let p(n) : 1 + 3 + 5 + ...+ (2n - 1) = n2

For n = 1

p(1) : 1 = 12

1 = 1

Let p(n) is true for n = k, so

p(k) : 1 + 3 + 5 +...+ (2k-1) = k2 ----(1)

we have to show that

1 + 3 + 5 +....+(2k - 1) +2(k + 1) -1 = (k + 1)2

Now,

{1+3+5+..+(2k - 1)} + (2k + 1)

= k2 + (2k + 1)                [Using equation (1)]

= k2 + 2k + 1

= (k + 1)2

⟹ p(n) is true for n = k + 1

⟹ p(n) is true for all n ϵ N by PMI

 

Mathematical Induction – Exercise 12.2 – Q.6

Mathematical Induction – Exercise 12.2 – Q.6

Put n = 1

⟹ p(n) is true for n = 1

Let p(n)  is true for n = k, so

We have to show that,

Now,

P(n)  is true for n = k  + 1

P(n) is true for all n ϵ N by PMI


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