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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.

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.

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.

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.

Python 2, 45 bytes

lambda a,b:(a+~b)*(a-b)*(3*(a+b)**2+a-b-2)/12

Not actually sure if that's the shortest, but thought it'd be worth posting anyway.

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.