<s>[INST] 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.
 [/INST] 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
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