Well, I can definitely help with the divisor. If the number of periods
is x, use the following:

FACT(x)

The FACT function returns the factorial, more commonly designated thus:

x!

I think that won't work -- Factorial would multiply, and for
a weighted average  divisor, it would have to add the weights.

Got this working for the divisor, though:

=COUNT(OFFSET($N10,0,0,-BK$2,1))*
(1+COUNT(OFFSET($N10,0,0,-BK$2,1)))/2

Now, just need to make the array {1;2;3;4;5;6;7;8;9;10;11;12}
part "flexible"...

Don't know if that can be done; have also tried writing in
VBA, and have posted as separate msg.

But this might be an easier formula  for your
divisor:

(x+1)/2*(x)