Formula to calculate $(1n)+(2(n-1))+(3(n-2)+(4(n-3)) ...$ How would I go about calculating something like $(1\times4)+(2\times3)+(3\times2)+(4\times1)$? It is the equivalent of $(4+3+2+1)+(3+2+1)+(2+1)+(1)$.

Yes, the two expressions you write are equivalent.  If you want to calculate that type of sum/product quickly for a similar longer expression, so you can find an explicit formula.  In your example, you go from 1 to 4, so let's call 4=n.
Then your expression $= \Sigma_{i=1}^n i(n-i+1) = \Sigma_{i=1}^n (i(n+1)- i^2) = (n+1) \Sigma_{i=1}^n i - \Sigma_{i=1}^n i^2 = (n+1) * n(n+1)/2 - n(n+1)(2n+1)/6 = n(n+1)(n+2)/6$
So when $n=4$, we get 4*5*6/6=20, which we know is correct.  But we can also plug in larger values of $n$ and get the answer instantly.
