I'm trying to find lower and upper bounds for the following two partial sums, respetively: $$ \sum_{k = 0}^r k \cdot {n \choose k} \cdot p^k (1 - p)^{n - k} \tag{1}\label{eq1} $$ and $$ \sum_{k = 0}^r k^2 \cdot {n \choose k} \cdot p^k (1 - p)^{n - k} \tag{2}\label{eq2} $$ where $r < n$ is fixed and $p > \frac{1}{2}$.
What I’ve tried. I have checked this answer which is similar to \eqref{eq1} in this question. To compute the bounds for \eqref{eq1}, we first rewrite it into $$ (1 - p)^n \cdot \sum_{k = 0}^r k \cdot {n \choose k} \cdot \mu^k \tag{3}\label{eq3} $$ where $\mu = \frac{p}{1 - p} > 1$. Then, we can apply the the given answer by considering the ratio $$ \frac{k{n \choose k} \mu^k}{r{n \choose r} \mu^r} = \frac{k}{r} \cdot \frac{(n - r)! r!}{(n - k)! k!} \cdot \mu^{k - r} = \frac{(n - r)! (r - 1)!}{(n - k)! (k - 1)!} \cdot \mu^{k - r}. \tag{4}\label{eq4} $$ But it is quite diffcult for me to bound \eqref{eq4} using Stirling's formula. I face the same difficulty in dealing with \eqref{eq2}. I have checked the paper cited in the answer but it did not help.
It would be appreciated if someone could give me any hint to bound the two partial sums.