Can you find a formula for this?

The different terms represent the number of dots in different diagonal slices of a regular tetrahedron, so the sum of the terms will be a tetrahedral number. The following picture shows the specific case for $n = 4$ :

Since the $n^{\text{th}}$ tetrahedral number is $\frac{n(n + 1)(n + 2)}{6}$ , that means $a = 0$ , $b = 1$ , $c = 2$ , $d = 6$ , and $a + b + c + d = \boxed{9}$ .

In summation notations, we have $\displaystyle \sum_{k=1}^n k = \dfrac {n(n+1)}2$ , $\displaystyle \sum_{k=1}^n k^2 = \dfrac {n(n+1)(2n+1)}6$ , and

$\begin{aligned} S & = 1(n) + 2(n-1) + 3(n-2)+ \cdots + (n-1)(2) + n(1) \\ & = \sum_{k=1}^n k(n+1-k) \\ & = \sum_{k=1}^n ((n+1)k-k^2) \\ & = (n+1) \sum_{k=1}^n k - \sum_{k=1}^n k^2 \\ & = \frac {(n+1)n(n+1)}2 - \frac {n(n+1)(2n+1)}6 \\ & = \frac {n(n+1)}6 \cdot (3(n+1) - (2n+1)) \\ & = \frac {n(n+1)(n+2)}6 \end{aligned}$

Therefore $a+b+c+d = 0+1+2+6 = \boxed 9$ .

$1 \times n + 2 \times (n-1) + 3 \times (n-2) + ... + (n-1) \times 2 + n \times 1$ can be rewritten as:

$1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ... + (1 + 2 + 3 + ... + n)$

The expression above has (n) 1's, (n-1) 2's, (n-2) 3's, ... , and 1 n'(s). Therefore it is equivalent to the sum of the first n triangle numbers. The formula for the x $^{th}$ triangle number is the first formula,

$1 + 2 + 3 + 4 + ... + x = \frac{1}{2}(x^2+x)$ , and so we plug in 1, 2, 3, all the way to n for x, and then sum it all up.

So the expression is

$\frac{1}{2}[(1^2 + 2^2 + 3^2 + ... + n^2) + (1 + 2 + 3 + ... + n)] = \frac{1}{2}[\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}]$ ,

and then through some simplification we arrive at the formula,

$1 \times n + 2 \times (n-1) + 3 \times (n-2) + ... + (n-1) \times 2 + n \times 1 = \frac{n(n+1)(n+2)}{6} \Rightarrow a = 0, b = 1, c = 2, d = 6$

$a + b + c + d = 0 + 1 + 2 + 6 = \boxed{9}$ .

The series can be expressed as

$T_r = r(n - r +1)$

$T_r = (n + 1)r - r^2$

$\sum_{r=1}^n T_r = (n+1)\sum_{r=1}^nr - \sum_{r=1}^n r^2$

$S = (n+1)\frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$

$S = \frac{n(n+1)(n+2)}{6}$

So, $a = 0$ , $b = 1$ , $c = 2$ and $d = 6$

So, $\boxed{a + b + c + d = 9}$