Given a function, find values of when it is equal to $100$. Question: Let $f(n) = 2n - [\frac{(1 + \sqrt{8n – 7})}{2}]$ for each positive integer, where $[x]$ is denoted for the largest integer less than or equal to $x$ for any real number $x$. Find the sum of all possible value(s) $n$ such that $f(n) = 100$.
My solution:
If $f(n) = 100$ then $2n - [\frac{1 + \sqrt{8n – 7}}{2}] = 100$. It is also known that $[x] = \lfloor{x}\rfloor$, so I will be using $\lfloor{x}\rfloor$ throughout the rest of my answer.
First of all, $2n - [\frac{1 + \sqrt{8n – 7}}{2}] = 2n - \lfloor(2 + \frac{\sqrt{8n – 7}}{2})\rfloor$.
So, since $2n - \lfloor(2 + \frac{\sqrt{8n – 7}}{2})\rfloor = 100$, and $\lfloor{x}\rfloor \approx x$, we want $2n - (2 + \frac{\sqrt{8n – 7}}{2}) \approx 100$
Solving this equation, we get $2 + \frac{\sqrt{8n - 7}}{2} \approx 100 - 2n$. So, $\sqrt{8n - 7} \approx 196 - 4n.$ Since $\sqrt{8n - 7}$ is positive, $196 - 4n$ is also positive. So, $n > 49$.
Testing values of $n$, we see that $n = 55$ works. If $n > 55$, then $f(n) > 100$. This is because $2n$ grows much bigger than $\sqrt{8n - 7}$ when $n > 55$.
Hence, there is only one value of $n$, which is $55$, so the answer is $\boxed{55}$.
Could anyone review my solution for a better solution or if there are any mistakes in my solution? Feel free to write a better answer in the answers section and thanks in advance!

$$2n- \left\lfloor \frac{1+\sqrt{8n-7}}2\right\rfloor= 100$$
$$2n-100 =\left\lfloor \frac{1+\sqrt{8n-7}}2\right\rfloor $$
$$  \frac{\sqrt{8n-7}-1}2\le 2n-100 \le  \frac{1+\sqrt{8n-7}}{2}$$
$$4n-200-1 \le \sqrt{8n-7} \le 4n-200 + 1$$
$$(4n-201)^2 \le 8n-7 \le (4n-199)^2$$
The first inequality is satisfied for $$46 \le n \le 55$$
and the second inequality is satisfied by $$n \le 45 \lor n \ge 55$$
$n=55$ is the only solution.
