Reverse Triangular Sum

Notice that in the sum, the number 12 occurs 12 times, the number 11 occurs 11 times, and so on.

Hence, the sum is simply equal to $12\cdot 12 + 11\cdot 11 + 10\cdot 10 + ... + 1\cdot 1$

$=12^2+11^2+10^2+...+1^2$

Using the formula $1^2+2^2+3^2+...+n^2=\dfrac{n(n+1)(2n+1)}{6}$ , this is equal to $\dfrac{12(13)(25)}{6}=\boxed{650}$ .