`1^2 + 3^2 + 5^2 + (2n - 1)^2 = (n(4n^2 - 1))/3` prove this using mathematical induction

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First check for `n=1`

`1^2=(1cdot(4cdot1^2-1))/3`

`1=1`

So for `n=1` the identity holds. Now assume that the identity holds for all natural numbers greater or equal to `n` i.e.

`1^2+3^2+5^2+cdots+(2k-1)^2=(k(4k^2-1))/3,` `k leq n` **(1)**

Let's now check for `k=n+1`

`1^2+3^2+5^2+cdots+(2n-1)^2+(2(n+1)-1)^2=`

Now by using assumption (1) we get

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

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

`(4n^3-n+12n^2+12n+3)/3=`

`((n+1)(4(n+1)^2-1))/3`

which we were supposed to get. Hence our proof is done.

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