## Python 2, 45 bytes
<!-- language: lang-py -->

    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 *n*th [square pyramidal number][1], and `t(n)` be the *n*th [triangular number][2]. 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.

  [1]: https://en.wikipedia.org/wiki/Square_pyramidal_number
  [2]: https://en.wikipedia.org/wiki/Triangular_number