**Given**

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

**Proof**

Let,

S = 1 + 3 + 5 + .. + (2n-5) + (2n-3) + (2n-1)

S is the summation of first n odd numbers.

2*S = ( 1 + 3 + 5 + .. + (2n-5) + (2n-3) + (2n-1) ) + ( 1 + 3 + 5 + .. + (2n-5) + (2n-3) + (2n-1) )

By moving around the terms in above equation, we obtain

2*S = (1 + (2n-1)) + (3 + (2n-3)) + (5 + (2n-5)) + ….. + ((2n-5) + 5) + ((2n-3) + 3) + ((2n-1) + 1)

2*S = 2n + 2n + 2n + … + 2n

We know there are n terms in S. So the above equation must also contain n terms.

2*S = n*(2n)

2*S = 2*n^{2}

S = n^{2}

We know that n ∈ N. Therefore, S is a perfect square. Hence proved.