Prove that the 1+3+5+…+(2n-1)=n^2 for every positive integer

Given

1 + 3 + 5 + 7 + ⋯ + (2n-1) = n² ∀ n ∈ N

Proof

We will prove the statement by mathematical induction.

STEP 1

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

LHS = 1 + 3 + 5 + 7 + ⋯ + (2n-1) = 1

RHS = n² = 1² = 1

The statement is true for n = 1.

STEP 2

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

1 + 3 + 5 + 7 + ⋯ + ( 2k – 1 ) = k²

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

We need to prove,

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

Starting from LHS,

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

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

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

Substituting k² in the above equation

LHS = k² + ( 2( k + 1 ) – 1 )

LHS = k² + 2k + 1

LHS = ( k + 1 )²

LHS = RHS

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

Hence Proved.