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Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left(1+{\cal O}(n^{-1/2})\right),$$ from which $$\lim_{n\rightarrow\infty} \frac{\sum_{k=0}^n k^\alpha {n\choose k}}{n^\alpha 2^{n}}=2^{-\alpha}.$$ Note that this large-$n$ asymptotics is actually exact for all $n$ for $\alpha=1$.

Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left(1+{\cal O}(n^{-1/2})\right),$$ from which $$\lim_{n\rightarrow\infty} \frac{\sum_{k=0}^n k^\alpha {n\choose k}}{n^\alpha 2^{n}}=2^{-\alpha}\sum_{k=0}^n 2^{-n} {n\choose k}=2^{-\alpha}.$$ Note that this large-$n$ asymptotics is actually exact for all $n$ for $\alpha=1$.

Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left(1+{\cal O}(n^{-1/2})\right),$$ from which $$\lim_{n\rightarrow\infty} \frac{\sum_{k=0}^n k^\alpha {n\choose k}}{n^\alpha 2^{n}}=2^{-\alpha}.$$

Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left(1+{\cal O}(n^{-1/2})\right),$$ from which $$\lim_{n\rightarrow\infty} \frac{\sum_{k=0}^n k^\alpha {n\choose k}}{n^\alpha 2^{n}}=2^{-\alpha}.$$ Note that this large-$n$ asymptotics is actually exact for all $n$ for $\alpha=1$.

Expansion of the binomial around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left(1+{\cal O}(n^{-1/2})\right),$$ from which $$\lim_{n\rightarrow\infty} \frac{\sum_{k=0}^n k^\alpha {n\choose k}}{n^\alpha 2^{n}}=2^{-\alpha}.$$

Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left(1+{\cal O}(n^{-1/2})\right),$$ from which $$\lim_{n\rightarrow\infty} \frac{\sum_{k=0}^n k^\alpha {n\choose k}}{n^\alpha 2^{n}}=2^{-\alpha}.$$