Correct option is (D) \(\frac{n(4n^2-1)}{3}\)
\(S_n=1^2+3^2+5^2+......+\) upto n terms
\(=1^2+3^2+5^2+......+(2n-1)^2\)
\(=\sum(2n-1)^2\)
\(=\sum(4n^2-4n+1)\)
\(=4\sum n^2-4\sum n+\sum1\)
\(=\frac{4n(n+1)(2n+1)}6-\frac{4n(n+1)}2+n\)
\(=n\left(\frac23(2n^2+3n+1)-2(n+1)+1\right)\)
\(=\frac n3(4n^2+6n+2-6n-6+3)\)
\(=\frac n3(4n^2-1)\)