Prove by the principle of mathematical induction:1 + 3 + 5 + … + (2n

1.	Prove by the principle of mathematical induction:1 + 3 + 5 + … + (2n – 1) = n2 i.e., the sum of first n odd natural numbers is n2.
Answer» Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n² Now, let us check for the P (n) is true for n = 1 P (1) = 1 =1² 1 = 1 P (n) is true for n = 1 Then, let’s us check for the P (n) is true for n = k P (k) = 1 + 3 + 5 + … + (2k – 1) = k² … (i) Now, we have to show that 1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)² Then, 1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = k² + (2k + 1) = k² + 2k + 1 = (k + 1)² P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction:1 + 3 + 5 + … + (2n – 1) = n2 i.e., the sum of first n odd natural numbers is n2.

Answer»

Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n²

Now, let us check for the P (n) is true for n = 1

P (1) = 1 =1²

1 = 1

P (n) is true for n = 1

Then, let’s us check for the P (n) is true for n = k

P (k) = 1 + 3 + 5 + … + (2k – 1) = k² … (i)

Now, we have to show that

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)²

Then,

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1

= k² + (2k + 1)

= k² + 2k + 1

= (k + 1)²

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.