Solve summation $\sum_{i=1}^n \lfloor e\cdot i \rfloor $ 
How to solve $$\sum_{i=1}^n \lfloor e\cdot i \rfloor $$ 
  For a given $n$. 

For example, if $n=3$, then the answer is $15$, and it's doable by hand. But for larger $n$ (Such as $10^{1000}$) it gets complicated . Is there a way to calculate this summation?
 A: Let's approximate $e$ by a rational number $A$ such that we have a positive integer $k$ such that $10^kA$ is an integer (e.g., $A=2.71$ with $k=2$).
We can then use the fact that $\lfloor (m\cdot10^k+i)A \rfloor = m\cdot 10^k A+\lfloor Ai \rfloor$ for any positive integer $m$.
Then, for a positive integer $r$,
$$ \large
\sum_{i=1}^{r \cdot 10^k} \lfloor Ai \rfloor = r \sum_{i=1}^{10^k} \lfloor Ai \rfloor +\frac{r(r-1)}{2}10^{2k}A.$$
Since the sum is a non-decreasing function of $A$, we can bound the desired sum with two 
rational approximations of $e$, one large and one smaller.  
For instance, with $A=2.71828$ and $A=2.71829$ (and $k=5$, $r=10^4$) we can find that 
$$
1359140000859160000 < \sum_{i=1}^{10^9} \lfloor ei \rfloor < 1359145000859150000
$$
With $A=2.7182818$ and $A=2.7182819$ (and $k=7$, $r=10^2$) we find that 
$$
1359140900859141000 <\sum_{i=1}^{10^9} \lfloor ei \rfloor < 1359140950859140900.$$
By using better approximations, we can get tighter bounds.
A: disclaimer: This is not a complete answer, just an approximation I came up with that was too long to comment.
If we truncate $e $ at $3$ decimal places (so it is 2.718), we can get a decent approximation:
$$\sum_{i=1}^n \lfloor 2.718i \rfloor \approx \frac32 (n(n+1) - \lfloor \frac{n}3 \rfloor ( \lfloor \frac{n}3 \rfloor + 1)) + \frac12 \lfloor \frac{n}4 \rfloor ( \lfloor \frac{n}4 \rfloor + 1) = f(n)$$
I used the program provided in the comments to compare the answers for $n = 542$ (I know, not very large, but the program timed out for me there).
$$\sum_{k=1}^{542} \lfloor ei \rfloor = 399,732$$
$$f(542) = 401,769$$
So that the error is roughly $0.51$%
From here, you could numerically find an approximation for the error as a function of $n$, say $\delta(n)$, and a much better approximation for large values would be $f(n) - \delta(n) $ 
You would never find an exact answer this way, but if you're only curious in the value the sum takes for large $n$, this could be a decent way to get a nice approximation (it could be hard for a code to compute a sums exact value at very large $n $)
