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I am having some trouble understanding how part of this formula is derived.

Taken from:
http://jonisalonen.com/2013/deriving-welfords-method-for-computing-variance/

$(x_N−barx_N)^2+sum_i=1^N−1(x_i−barx_N+x_i−barx_N−1)(barx_N−1–barx_N)$
$=(x_N−barx_N)^2+(barx_N–x_N)(barx_N−1–barx_N)$

I can't seem to derive the right side from the left.

Any help explaining this would be greatly appreciated! Thanks.

I think the key is understanding:

and also:

The above reduces to:

This is a good post: https://alessior.wordpress.com/2017/10/09/onlinerecursive-variance-calculation-welfords-method/

The key to understanding is the following algebraic identity:
$$sum_i=1^N (x_i − barx_N) = 0$$
which basically says that the algebraic sum of deviations from the mean is zero. It is quite straightforward to derive this from the definition of mean:

$$ barx_N = frac1N sum_i=1^N x_i$$

This can be rewritten as:
$$ N barx_N = sum_i=1^N x_i$$

Since mean ($barx_N $) is a constant you can rewrite multiplying it by $N$ as adding it $N$ times:
$$ implies sum_i=1^N barx_N = sum_i=1^N x_i $$

Which reduces to:
$$sum_i=1^N (x_i − barx_N) = 0$$

Now let us look at the summation on the LHS
$$sum_i=1^N−1(x_i−barx_N + x_i−barx_N−1)
= sum_i=1^N−1((x_i−barx_N) + (x_i−barx_N−1)) \
= sum_i=1^N−1(x_i−barx_N) +sum_i=1^N−1 (x_i−barx_N−1)$$

Now apply the identity stated in the beginning to the above equation, the second term vanishes on the RHS.
$$sum_i=1^N−1(x_i−barx_N) +sum_i=1^N−1 (x_i−barx_N−1)
= sum_i=1^N−1(x_i−barx_N) + 0 $$

We just need a little more algebraic manipulation for the first term on the RHS.
We need the index $i$ to go from $1$ to $N$.

beginalign
sum_i=1^N−1(x_i−barx_N)
&= left(sum_i=1^N-1(x_i−barx_N) right) + (x_N −barx_N) - (x_N −barx_N) \
&= left(sum_i=1^N-1(x_i−barx_N) + (x_N −barx_N) right) - (x_N −barx_N) \
&= left(sum_i=1^N(x_i−barx_N) right) - (x_N −barx_N)
endalign

The first term on the LHS vanishes leading to:
$$sum_i=1^N−1(x_i−barx_N) = (barx_N − x_N) $$

We have now derived
$$sum_i=1^N−1(x_i−barx_N + x_i−barx_N−1) = (barx_N − x_N) $$

and plug this into the following equation:

beginalign
(x_N−barx_N)^2 + sum_i=1^N−1(x_i−barx_N+x_i−barx_N−1)(barx_N−1–barx_N)
&= (x_N−barx_N)^2 + (barx_N−1–barx_N) sum_i=1^N−1(x_i−barx_N+x_i−barx_N−1) \
&= (x_N−barx_N)^2 + (barx_N−1–barx_N) (barx_N − x_N)
endalign

which completes the derivation.

One can simplify this expression further:

beginalign
(x_N−barx_N)^2 + (barx_N−1–barx_N) (barx_N − x_N)
&= (x_N−barx_N) left [ (x_N−barx_N) - (barx_N−1–barx_N) right ] \
&= (x_N−barx_N) (x_N − barx_N−1)
endalign