**Given**

1 + 3 + 5 + 7 + ⋯ + (2n-1) = n^{2} ∀ n ∈ N

**Proof**

We will prove the statement using mathematical induction.

**STEP 1**

In this step, we check if the statement is true for n = 1.

1 + 3 + 5 + 7 + ⋯ + (2n-1) = 1

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

LHS = RHS

The statement is true for n = 1.

**STEP 1**

Let the statement be true for n = k ∀ k ∈ N.

1 + 3 + 5 + 7 + ⋯ + ( 2k – 1 ) = k^{2}

Now, we need to check if the statement is true for n = k + 1 given that it is true for n = k.

Basically we need to prove,

1 + 3 + 5 + 7 + ⋯ + ( 2( k + 1 ) – 1 ) = ( k + 1 )^{2}

We start from LHS

LHS = 1 + 3 + 5 + 7 + ⋯ + ( 2( k + 1 ) – 1 )

= 1 + 3 + 5 + 7 + ⋯ + ( 2k – 1 ) + ( 2( k + 1 ) – 1 )

= ( 1 + 3 + 5 + 7 + ⋯ + ( 2k – 1 ) ) + ( 2( k + 1 ) – 1 )

Substituting k^{2} in the above equation

= k^{2} + ( 2( k + 1 ) – 1 )

= k^{2} + 2k + 1

= ( k + 1 )^{2}

= RHS

Therefore, we can say that the given statement is true for n = k + 1 whenever it is true for n = k ∀ k ∈ N. By principle of mathematical induction the given statement is true for ∀ n ∈ N.

Hence Proved.