We have
an = (2n−1)2 = 4n2−4n+1
Hence, the sum of n terms is
Sn = 4[

n(n+1)(2n+1)

6

] - 4[

n(n+1)

2

] + n
=

n(n+1)

2

[

4(2n+1)

3

−4] + n
=

n(n+1)

2

[

8n+4−12

3

]+n
=

n(n+1)(8n−8)

6

+n =

n(n+1)(8n−8)+6n

6

=

n[8n2−8n+8n−8+6]

6

=

n[8n2−2]

6

=

n(4n2−1)

3

=

n

3

(2n + 1) (2n - 1)