For demonstrating the identity, we'll use the method of mathematical induction, which consists in 3 steps:

1) verify that the method works for the number 1;

2) assume that the method works for an arbitrary number, k;

3) prove that if the method works for an arbitrary number k, then it work for the number k+1, too.

4) after the 3 steps were completed, then the formula works for any number.

Now, we'll start the first step:

1) 1^2=1*(4*1^2-1)/3 => 1=3/3=1 true.

2) 1^2 + 2^2 + ... + (2k-1)^2 = k(4k^2-1)/3 , true.

3) If 1^2 + 2^2 + ... + (2k-1)^2 = k(4k^2-1)/3, then

1^2 + 2^2 + ... +(2k-1)^2 + (2k+1)^2=(k+1)(4(k+1)^2-1)/3

Let's see if it is true.

For the beginning, we notice that the sum from the left contains the assumed true equality,1^2 + 2^2 + ... + (2k-1)^2=k(4k^2-1)/3. So, we'll re-write the sum by substituting a part of it with k(4k^2-1)/3.

k(4k^2-1)/3 + (2k+1)^2 = (k+1)(4(k+1)^2-1)/3

4k^3-k+3(4k^2+4k+1)=(k+1)(4k^2+8k+4-1)

4k^3-k+12k^2+12k+3=4k^3+8k^2+3k+4k^2+8k+3

4k^3+12k^2+11k+3=4k^3+12k^2+11k+3 true.

4) The 3 steps were completed, so the identity is tru for any value of n.

1^2 + 2^2 + ... + (2n-1)^2 = n(4n^2-1)/3

The first answer gives another result and the identity is not demonstrated.

I opened the paranthesis for n(2n-1)(4n-1)/3=n(8n^2-6n+1)/3 and it is not equal to n(4n^2-1)/3.

To calculate 1^2+2^2+3^2+....(2n-1)^2.

The number of terms is 2n-1 .

To sum the above we use the formula 1^2+2^2+3^2*....x^2 = x(x+1)(2x+1)/6, where x is any natural number.

Substituting, for x =2n-1, we get

So S2n-1 = (2n-1)(2n-1+1)[2(2n-1)+1]/6

=(2n-1)(2n)(4n-1)/6

= n(2n-1)(4n-1)/3. Therefore

1^2+2^2+3^3+4^2+....+(2n-1) = n(2n-1)(4n-1)/3