Generalizing a series of numbers I know that I studied this long ago but I can't seem to bring the information to mind.  I am looking to construct a general formula for the following sequence of numbers ->
when $x=4$ : $n=(3+x)+2(x-1)$;
when $x=5$ : $n=(3+x)+3(x-1)+2(x-2)+1(x-3)$;
when $x=6$ : $n=(3+x)+4(x-1)+3(x-2)+2(x-3)+1(x-4)$;
etc.  I worked out the pattern but can't generalize this into a formula.  Can anyone help?

Note that we can rewrite your statements:
when $x=4:n=(3+x)+2(x-1)$. Obviously, if $x=4$, then we have $n=(3+4)+2(4-1) = 13$.  
$\begin{align} x=5:n&=(3+x)+3(x-1)+2(x-2)+1(x-3)\\ &= 3+4+3(4-1)+2(4-2)+1(4-3)\\ &= 7+9+4+1 \\ &= 21\end{align}$  
$\begin{align} x=6:n&=(3+x)+4(x-1)+3(x-2)+2(x-3)+1(x-4) \\ &= 3+6+4(6-1)+3(6-2)+2(6-3)+1(6-4) \\ &=9+20+12+6+2 \\ &= 49 \end{align}$
I searched for a sequence containing $13,21,49$ in the Online Encylopedia of Integer Sequences, but I could not find anything that seems particularly helpful. So I'm not sure if there is a particular sequence here.
You could rewrite all integers in terms of $x$:
$$\begin{align}x=4:n &= 1(3+x)+(x-2)(x-1) \\ &= 3+x+x^2-3x-2 \\ &= x^2-2x+1 \\ &= (x-1)(x-1)\end{align}$$
$$\begin{align} x=5: n &= 1(3+x)+(x-2)(x-1)+(x-3)(x-2) + (x-4)(x-3)\\ &= 3+x+x^2-3x-2+x^2-5x+6+x^2-7x+12 \\ &= 3x^2-14x+1 \end{align}$$
You could use $\ldots$ notation to make the process easier: e.g. $(x-5)(x-4) + \ldots +(x-1)$ will help write the longer expressions more compactly.
I know this is not a complete answer, but I hope it helps or gives you some useful ideas.
