watching a proof for the sums of n number we have we have that:
$$(n+1)^2 - n^2 = 2n +1$$
how does that latter equation becomes
$$(n+1)^2 - 1^2 =2\sum_{k=1}^{n}k + n $$
I know there are better proofs but I would like to see the explanation of this one because if I understand this I can derive the next proof of sums of squares
I'd handle it as a telescoping sum.
Or if you're not yet comfortable with that, write it out $n$ times, for $n, n-1, n-2, ... 1$:
$$ \begin{align} (n+1)^2 - (n-0)^2 &= 2(n-0) +1\\ (n-0)^2 - (n-1)^2 &= 2(n-1) +1\\ (n-1)^2 - (n-2)^2 &= 2(n-2) +1\\ (n-2)^2 - (n-3)^2 &= 2(n-3) +1\\ (n-3)^2 - (n-4)^2 &= 2(n-4) +1\\ .\\ .\\ .\\ 2^2 -1^2 &= 2(1) +1 \end{align}$$
and then add them all up and enjoy all the cancellations that happen on the left hand side.
(This is really just what a telescoping sum is, but writing out all of the details.)