Demonstrate the equality: 1^2 + 2^2 + ... + (2n-1)^2 = n(4n^2-1)/3

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