Evaluation of the sum $\sum_{i=1}^{\lfloor na \rfloor} \left \lfloor ia \right \rfloor $

Let $a$ be a positive proper fraction and $n$ is any integer then evaluate the following sum, $$\sum_{i=1}^{\left \lfloor na \right \rfloor\atop} \left \lfloor ia \right \rfloor $$

I think that probably some counting argument will give us a solution of the problem but I can't find it. Any help is appreciated.

Actually the question aroused due to my efforts in trying to find out a simple formula for calculating the Legendre Symbol. Following is my approach to finding a closed form for the sum $$\sum_{i=1}^{\frac{p-1}{2}} \left \lfloor \dfrac{ia}{p} \right \rfloor $$ where $\operatorname{gcd}(a,p)=1$ and $a$ is odd.

My Try

I have tried to approach the above problem in the following manner. Consider a rectangle in the Cartesian Plane with coordinates of the vertices $A(0,0)$, $B(\dfrac{p}{2},0)$, $C(\dfrac{p}{2},\dfrac{a}{2})$ and $D(0,\dfrac{a}{2})$.

Now draw the two diagonals $AC$ and $BD$. Let them intersect at $O$. Denote the number of lattice points in region $R$ with no lattice point (if any) on the boundaries is counted by $\Lambda(R)$.

Then using this notation we have, $$\Lambda(\triangle AOB)+\Lambda(\triangle BOC)+\Lambda(\triangle COD)+\Lambda(\triangle DOA)=\Lambda(\square ABCD)$$

But since $\triangle AOB \equiv \triangle COD$ and $\triangle BOC \equiv \triangle DOA$, we get $\Lambda(\triangle AOB)=\Lambda(\triangle COD)$ and $\Lambda(\triangle BOC)=\Lambda(\triangle DOA)$

Therefor our expression reduces to, $$2(\Lambda(\triangle AOB)+\Lambda(\triangle BOC)=\Lambda(\square ABCD) \implies 2\Lambda(\triangle ABC)=\Lambda(\square ABCD)$$

But, $$\Lambda(\triangle ABC)=\sum_{i=1}^{\frac{p-1}{2}} \left \lfloor \dfrac{ia}{p} \right \rfloor $$ and $$\Lambda(\square ABCD)=\left(\dfrac{p-1}{2} \right)\left \lfloor \dfrac{a}{2} \right \rfloor$$ hence we have, $$2\sum_{i=1}^{\frac{p-1}{2}} \left \lfloor \dfrac{ia}{p} \right \rfloor=\left(\dfrac{p-1}{2} \right)\left(\dfrac{a-1}{2} \right )$$

But from this I can't interpret the given sum. What is the mistake? There must be a mistake because the parity of both side of the inequality doesn't hold if $p \equiv a \equiv 3$ $(\operatorname{mod}4)$. Can anyone help me in this respect?

Let $a = \displaystyle\frac{p}{q}$ where $(p,q) = 1$ and $p < q$. Then:

$$\sum_{i=1}^{\lfloor na\rfloor} \lfloor ia \rfloor = \sum_{i=1}^{\lfloor na\rfloor} ia - \sum_{i=1}^{\lfloor na\rfloor} \lbrace ia \rbrace $$

Where $\lbrace x\rbrace$ is the fractional part of $x$.

Note that:

$$\sum_{i=1}^{\lfloor na\rfloor} \lbrace ia \rbrace = \frac{1}{q}\sum_{i=1}^{\lfloor na\rfloor} ip\text{ mod }q $$

We can get some interesting results from that depending on $n \text{ mod }q$.

Not quite an answer but I couln't get this to look correct as comment. But I think $$ 0 <\sum_{i=1}^{\left \lfloor na \right \rfloor} \left \lfloor ia \right \rfloor < a\cdot {\left \lfloor na \right \rfloor}\frac{{\left \lfloor na \right \rfloor} + 1}{2} $$

Should be true.