How to solve $\sum_{i=a}^b\sum_{j=i}^c(j-1)$

Question

I really have looked for about 2 hours on the Internet, but I cannot find how to solve $\sum_{i=a}^b\sum_{j=i}^c(j-1)$. Can somebody please help?

Answer 1

Hint: $$\sum_{i=1}^{n}(a-b)=n\cdot (a-b)$$

i.e. to say, you have to watch out if the expression contained inside the sigma operator does at all depend on the variable ($i$ in the above case)

Hint 2:

The inner summation varies with $j$, but the expression contained in it $(i-1)$ is independent of $j$.

Hint 3:

The inner sum easily reduces to $(c-i+1)\cdot(i-1)$