1. ## Series Problem

An and Bn are two series given by:
An = $1^2$+ $5^2$ + $9^2$+ $13^2$ + …… + (4n–3 $)^2$
Bn = $3^2$ + $7^2$ + $11^2$ + $15^2$ + ….
For n = 1, 2, 3,….
i. Find the nth term of Bn.
Tn = $(4n - 1)^2$

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

i have so far for this only
Sn = $1^2$+ $5^2$ + $9^2$+ $13^2$ + …… + (4n–3 $)^2$
- ( $3^2$ + $7^2$ + $11^2$ + $15^2$ + $(4n - 1)^2$)

and have absolutely no idea how to proceed from here

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

any help greatly appreciated
thanks

2. Originally Posted by jacs
An and Bn are two series given by:
An = $1^2$+ $5^2$ + $9^2$+ $13^2$ + …… + (4n–3 $)^2$
Bn = $3^2$ + $7^2$ + $11^2$ + $15^2$ + ….
For n = 1, 2, 3,….
i. Find the nth term of Bn.
Tn = $(4n - 1)^2$

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

$A_n=\sum_{r=1}^n (4r-3)^2=\sum_{r=1}^n \left[16r^2-24r+9\right]$

You should know the formula for the sum of the first $n$ integers and their squares.

Same for $B_n$

CB

3. Thanks CB
went ahead with the sigma and tried this, although i must be doing something wrong because i cannot get the required result. (pls see attached image)

4. Originally Posted by jacs
Thanks CB
went ahead with the sigma and tried this, although i must be doing something wrong because i cannot get the required result. (pls see attached image)
I agree, the given answer is $S_n$ not $S_{2n}$ . To show this compute $S_1=A_1-B_1$ and $S_2=A_2-B_2$.
CB