Ytukyg

Binomial expansion gives ${2n choose 0}cdot2+ {2n choose 1}cdot 2 +...+{2n choose n-1}cdot2+{2n choose n}=2^{2n}$. My question is to find the asymptotic order of $$F(n):={2n choose 0}sqrt{n}+ {2n choose 1}sqrt{n-1} +...+{2n choose n}sqrt{0}$$
as $nrightarrow infty$. In particular, is it true that for all $n$, $F(n)le Ccdot 2^{2n}$ for some constant $C$?

In particular, is it true that for all $n$, $F(n)le Ccdot 2^{2n}$ for some constant $C$?

No.

Let's parameterise the sum slightly non-intuitively (it's in the reverse order to the way you've written it): $$sum_{i=0}^n binom{2n}{n-i} sqrt i$$

Then by Stirling's approximation the $i$th term is approximately $$frac{1}{sqrt {2pi}} frac{(2n)^{2n+1/2}}{(n-i)^{n-i+1/2}(n+i)^{n+i+1/2}} sqrt i$$ and when $i = o(n)$ that in turn approximates to $$frac{2^{2n}}{sqrtpi} sqrtfrac{i}{n}$$

Since there are $sqrt{n}$ values of $i$ in the range $[frac12sqrt n, frac32sqrt n)$ you should not expect an upper bound which is asymptotically better than $$2^{2n} n^{1/4}$$

This may not be tight, but it's enough to answer the direct question.

Here is a first-term asymptotic analysis that gets the same result as Peter Taylor above, but with an explicit constant. Numerically, about 3 digits agreement are obtained for n ~ 1000.

$$sum_{k=0}^n binom{2n}{n-k}sqrt{k} sim binom{2n}{n}sum_{k=0}^nexp{(-k^2/n)}sqrt{k}=binom{2n}{n}n^{3/2},frac{1}{n}sum_{k=0}^nexp{(-nbig(frac{k}{n}big)^2)}sqrt{frac{k}{n}}.$$

In the 1st step the well-known expansion for the central binomial coefficients have been used and the 2nd I'm setting up for the interpretation of the sum as a Riemann integral as n $to infty$, i.e.,

$$sum_{k=0}^n binom{2n}{n-k}sqrt{k} sim binom{2n}{n}n^{3/2},
int_0^1exp{(-n,x^2)}sqrt{x},dx .$$
The standard asymptotic trick of increasing the upper integral limit go to $infty$ allows us to explicitly calculate the integral with the gamma function. Using the Stirling approximation for the binomial and collecting results the answer so obtained is

$$ sum_{k=0}^n binom{2n}{n-k}sqrt{k} sim frac{Gamma(3/4)}{2sqrt{pi}}
n^{1/4}2^{2n}$$

Too long for a comment.

Having fun playing with small numbers, I computed the values of

$$S_n=sum_{i=0}^n binom{2n}{n-i} sqrt i$$ for $100 leq n leq 10000$ (step size $= 100$). For $n=1000$, the value is almost $1.375times 10^{6021}$ (!!).

Performing a linear regression
$$log(S_n)=a+ b n + c log(n)$$
$$begin{array}{clclclclc}
text{} & text{Estimate} & text{Standard Error} & text{Confidence Interval} \
a & -1.08076 & 0.00069 & {-1.08213,-1.07939} \
b & +1.38629 & approx 0 & {+1.38629,+1.38629} \
c & +0.25239 & 0.00010 & {+0.25219,+0.25259} \
end{array}$$ while using skbmoore's answer, the coefficients would be $ {-1.06223,1.38629,0.25000}$.