1²+3²+5²….+(2n-1)²=1/3 n(4n²-1) by induction

Check for n=1...(True)
Let it be true for n=k
i.e.
1²+3²+5²….+(2k-1)²= k(4k²-1)/3 ...(i)

To prove: It is true for n=k+1

Soln.
1²+3²+5²….+[2(k+1)-1]²
= k(4k²-1)/3+ [2k+1]² ...(from i)
= k(2k+1)(2k-1)/3+ (2k+1)²
=(2k+1)[k(2k-1)/3-(2k+1)]
=(2k+1)(2k+3)(k+1)/3
=(k+1)[2(k+1)+1][2(k+1)-1]/3
=(k+1)[4(k+1)²-1]/3