153099

For the first sequence, I decided to do a table. I found out that the formula was 8n.
For the second sequence, I worked out that:
the first pattern was 3^2-1
the second pattern was 5^2-1
the third pattern was 7^2-1
the forth pattern was 9^2-1
With this and using trial and error, I was able to form a formula which was: (2n+1)^2-1

For the third sequence, I did a table like the one I did for the first sequence. I found out that the formula 6n+2.
For the forth sequence I thought I would try and find out how to get the vertical and horizontal sides and time them together, then minus n. The formula I got was (n+2)x(2n+1)-n.