I came across this fun pattern, can anyone give me a proof for this? The Pattern
$\begin{align*}1(8)&=(3^2)-1\\
2(8)&= (3+1)^2 \\
3(8)&= (3+2)^2-1\\
4(8)&= (3+3)^2 - 4\\
5(8)&= (3+4)^2- 9\\
6(8)&= (3+5)^2-16\end{align*}$
Conjecture
I think the pattern is that the numbers appear in the following form:
$$8(n+1) = (3+n)^2 - (n-1)^2$$
Please correct me if I'm wrong, I'm not really that good at math.

If you are just learning algebra, this might not be obvious.  But, the rule to keep in mind is that multiplication distributes over addition.  On the left-hand side, you have $8(n+1)$ which after distribution equals $8n+8.$  Now we have to show that the right-hand side gives the same thing.
Let's work out the two terms one at a time.
$(n+3)^2 = (n+3)(n+3) = n(n+3) + 3(n+3) = n^2 + 3n + 3n+9 = 
n^2+6n+9$
$(n-1)^2 = (n-1)(n-1) = n(n-1) + (-1)(n-3) = n^2  - n - n +1 = n^2-2n+1$
$(n+3)^2 - (n-1)^2 = (n^2+3n+9)-(n^2-2n+1) = n^2+6n+9 - n^2+2n-1 = 8n+8$
As required.
Hope this helps
