Check asymptotic of
C = $\sum_{k = 0}^{\frac{n}{2} - \sqrt{n}} k \binom{n}{k} = f(n) + O(g(n))$
In the beginning I tried to simplify the expression under the sum: $$k\binom{n}{k} = k \frac{n!}{k!(n-k)!} = \frac{n!}{(k - 1)!(n - k)!} = n \frac{(n-1)!}{(k - 1)!((n-1) - (k-1))!} = n \binom{k-1}{n-1}$$ After I wrote out the resulting expression: $$\sum_{k = 0}^{\frac{n}{2} - \sqrt{n}} k \binom{n}{k} = \sum_{k = 1}^{\frac{n}{2} - \sqrt{n}} k \binom{n}{k} = \sum_{k = 1}^{\frac{n}{2} - \sqrt{n}} n \binom{n - 1}{k -1} = n \sum_{k = 1}^{\frac{n}{2} - \sqrt{n}} \binom{n - 1}{k -1}$$
At this point, I do not know what to do next. At first, I thought about expressing something through the formula $(1+x)^n$ taking derivatives and finding functions. I also tried to get this amount through the binomial of Newton, but failed.
The summand has a peak at $(n - 1)/2$, where for large $n$ and constant $\xi \in (-1, 1)$ it can be approximated as $$\binom {n - 1} {\frac {n - 1} 2 + \frac {n \xi} 2} \sim \frac {2^n} {\sqrt {2 \pi n}} e^{n \phi(\xi)}, \\ \phi(\xi) = -\frac 1 2 (1 - \xi) \ln(1 - \xi) - \frac 1 2 (1 + \xi) \ln(1 + \xi).$$ The maximum of $\phi$ is located at zero, where $\phi(\xi) \sim -\xi^2/2$.
$f$ can be derived by proving that the sum is asymptotically equivalent to the corresponding integral. The integral is then amenable to Laplace's method (a variation of the method where the critical point is asymptotically close to one of the endpoints): $$n \sum_{k = 0}^{n/2 - \sqrt n - 1} \binom {n - 1} k \sim \frac {n^2} 2 \frac {2^n} {\sqrt {2 \pi n}} \int_{-\infty}^{-2/\sqrt n} e^{-n \xi^2/2} d\xi = \operatorname{erfc}(\sqrt 2) n \hspace{1.5px} 2^{n - 2}.$$