Python 2, 45 bytes
lambda a,b:(a+~b)*(a-b)*(3*(a+b)**2+a-b-2)/12
Closed form solution - not the shortest, but I thought it'd be worth posting anyway.
Explanation
Let p(n) be the nth square pyramidal number, and t(n) be the nth triangular number. Then, for n over the range a, ..., b:
- ∑n =
t(b)-t(a-1), and - ∑n² =
p(b) - p(a-1) - So (∑n)²-∑n² =
(t(b)-t(a-1))² - (p(b) - p(a-1)).
This expression reduces to that in the code.
