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.