For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
By dividing the numbers between $1$ and $5^m$ as intervals of $5^k$, I was getting the following expression:
$$\binom{5^{m-k}}{2}$$ which is not only too good to be true but it turns out it is wrong. Any suggestions on how I should approach this?
Edit: After reading through @heropup's example, I am starting to realize that I may have forgotten $5^k$ term and so the following could be correct.
$$5^k \binom{5^{m-k}}{2}$$
Does that sound correct?