A question of approximation ask to prove that if $f$ is continuous on $[0,1]$ and $f(0)=f(1)=0$, then $$\sum_{k=1}^{n-1} \left[ \binom nk f\left(\dfrac kn\right)\right] x^k(1-x)^{n-k}$$ converges to $f$ uniformly as $n\to \infty$. Note that this question is not the same as Bernstein polynomial, the square bracket $[]$ stands for floor function. Assume we already know the Bernstein polynomials $$B_n(x)=\sum_{k=1}^{n-1}\binom nk f\left(\dfrac kn\right) x^k(1-x)^{n-k}$$converges to $f(x)$ uniformly on $[0,1]$, then it suffices to show their difference $$\sum_{k=1}^{n-1} \left\{\binom nk f\left(\dfrac kn\right)\right\} x^k(1-x)^{n-k}$$ which is bounded by $D_n(x)=\sum_{k=1}^{n-1} x^k(1-x)^{n-k}$, converges uniformly to $0$.
The simplified expression for $D_n(x)$ is $$D_n(x)=\dfrac{x(1-x)((1-x)^{n-1}-x^{n-1})}{1-2x}$$ where $D_n(1/2)=(n-1)/4^n$. By plotting graph, it seems $D_n(x)$ does converges uniformly to $0$, but I couldn't think of a method to prove this. My idea was to prove $f$ has maximum near $x=1/n$ and $x=1-1/n$ from observation, and bound the whole function from these two points. It is hard to solve $D_n'(x)=0$ directly, so I think there might be some methods which doesn't require derivative.
For all $0\leq x \leq 1$ and $n\geq 1$, $$ \sum\limits_{k = 1}^{n - 1} {x^k (1 - x)^{n - k} } \le \sum\limits_{k = 1}^{n - 1} {\frac{1}{n}\binom{n}{k}x^k (1 - x)^{n - k} } \le \frac{1}{n}\sum\limits_{k = 0}^n {\binom{n}{k}x^k (1 - x)^{n - k} } = \frac{1}{n}. $$