An and Bn are two series given by:

An = $\displaystyle 1^2 $+ $\displaystyle 5^2$ + $\displaystyle 9^2 $+ $\displaystyle 13^2$ + …… + (4n–3$\displaystyle )^2$

Bn = $\displaystyle 3^2$ + $\displaystyle 7^2$ + $\displaystyle 11^2$ + $\displaystyle 15^2$ + ….

For n = 1, 2, 3,….

i. Find the nth term of Bn.

Tn = $\displaystyle (4n - 1)^2$

ii. If Sn = An – Bn, prove that S2n = $\displaystyle -8n^2$

i have so far for this only

Sn = $\displaystyle 1^2 $+ $\displaystyle 5^2$ + $\displaystyle 9^2 $+ $\displaystyle 13^2$ + …… + (4n–3$\displaystyle )^2$

- ($\displaystyle 3^2$ + $\displaystyle 7^2$ + $\displaystyle 11^2$ + $\displaystyle 15^2$ + $\displaystyle (4n - 1)^2$)

and have absolutely no idea how to proceed from here

iii. Hence, evaluate $\displaystyle 101^2$ – $\displaystyle 103^2$ + $\displaystyle 105^2$ – $\displaystyle 107^2$ + …… + $\displaystyle 1993^2$ - $\displaystyle 1995^2$

any help greatly appreciated

thanks