Proof/derivation of $\lim\limits_{n\to\infty}{\frac1{2^n}\sum\limits_{k=0}^n\binom{n}{k}\frac{an+bk}{cn+dk}}\stackrel?=\frac{2a+b}{2c+d}$? I just came up with the following identity while solving some combinatorial problem but not sure if it's correct. I've done some numerical computations and they coincide.
$$\lim_{n\to \infty}{\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}\frac{an+bk}{cn+dk}}\;\stackrel?=\;\frac{2a+b}{2c+d}$$
Here $a$, $b$, $c$, and $d$ are reals except that $c$ mustn't $0$ and $2c+d\neq0$. I wish I could explain how I came up with it, but I did nothing but comparing numbers with the answer then formulated the identity, and just did numerical computations.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{\lim_{n\to \infty}\,{1 \over 2^{n}}
\sum_{k = 0}^{n}{n \choose k}{an + bk \over cn + dk}} =
\lim_{n\to \infty}\,{1 \over 2^{n}}
\sum_{k = 0}^{n}{n \choose k}\pars{an + bk}\int_{0}^{1}t^{cn + dk - 1}\,\,\,\dd t
\\[3mm] = &\
\lim_{n\to \infty}\,{1 \over 2^{n}}\int_{0}^{1}t^{cn - 1}\bracks{%
an\sum_{k = 0}^{n}{n \choose k}\pars{t^{d}}^{k} +
b\sum_{k = 0}^{n}{n \choose k}k\pars{t^{d}}^{k}}\dd t
\end{align}

However,
$\ds{\sum_{k = 0}^{n}{n \choose k}\xi^{k} = \pars{1 + \xi}^{n}
\quad\imp\quad
\sum_{k = 0}^{n}{n \choose k}k\,\xi^{k} = n\xi\pars{1 + \xi}^{n - 1}}$.

Then,
\begin{align}
&\color{#f00}{\lim_{n\to \infty}\,{1 \over 2^{n}}
\sum_{k = 0}^{n}{n \choose k}{an + bk \over cn + dk}} =
\\[3mm] = &\
\lim_{n\to \infty}\,{1 \over 2^{n}}\int_{0}^{1}t^{cn - 1}\bracks{%
an\pars{1 + t^{d}}^{n} +
bn\,t^{d}\pars{1 + t^{d}}^{n - 1}}\dd t
\\[3mm] = &\
\color{#f00}{\lim_{n \to \infty}\braces{\vphantom{\LARGE A}%
{n \over 2^{n}}\bracks{\vphantom{\Large A}a\,\mathrm{f}_{n}\pars{cn,d} +
b\,\mathrm{f}_{n - 1}\pars{cn + d,d}}}}
\end{align}

$$
\mbox{where}\quad
\begin{array}{|c|}\hline\\
\ds{\quad\mathrm{f}_{n}\pars{\mu,\nu} \equiv
\int_{0}^{1}t^{\mu - 1}\pars{1 + t^{\nu}}^{n}\,\dd t\quad}
\\ \\ \hline
\end{array}
$$


Can you take it from here ?. Maybe, an asymptotic study of the integral could be somehow reasonable. Otherwise, the integral is related to hypergeometric functions.

A: I believe a probabilistic argument can be given here.
Let $X_1, X_2, \ldots $ be iid Bernoulli random variables with success probability $p$. Then by Strong Law of Large Numbers $$\bar{X}_n:=\frac{1}n\sum_{i=1}^n X_i \stackrel{a.s.}{\to} p$$
Also for any continuous function $g$ we have $g(\bar{X}_n) \stackrel{a.s.}{\to} g(p)$. Now if $g$ is such that there exist a random variable $Y$ with finite expectation such that $g(\bar{X}_n) \le Y$ almost surely. Then by DCT we have $E(g(\bar{X}_n)) \to g(p)$.
Now this has lot of applications.
Example 1: Take $p=\frac12$. $g(x)=\frac{a+bx}{c+dx}$. Make sure $g$ is continuous $[0,1]$. Then clearly $g$ is bounded. Hence by DCT
$$E(g(\bar{X}_n))=\sum_{i=1}^n \frac{\binom{n}{k}}{2^n}g\left(\frac{k}{n}\right) \to g(\frac12)=\frac{2a+b}{2c+d}$$
Example 2: Take $p=\frac12$,$g(x)=e^{-x^2}$. Clearly $g$ is continuous and bounded.
Hence
$$\frac1{2^n}\sum_{i=1}^n \binom{n}{k}e^{-k^2/n^2} \to e^{-1/4}$$
Similarly we can obtain plethora of such results.
A: The reason this is true is that the binomial coefficients are strongly concentrated around the mean, when $k \sim \frac{n}{2}$.  Using some standard concentration inequalities (Chernoff is strong, but Chebyshev's inequality sufficies too), you can show that for any constant $c > 0$, $2^{-n} \sum_{ k \in [0, (1/2 - c ) n ] \cup [(1/2 + c)n, n]} \binom{n}{k} \rightarrow 0$ as $n \rightarrow \infty$.
Hence in your limit, the only terms that survive are when $k \sim \frac{n}{2}$, in which case you can cancel the $n$ throughout and get the right-hand side.
For the limit to hold, you clearly need $2c + d \neq 0$.  You would also need $c \neq 0$, as otherwise the $k = 0$ term of the sum will present difficulties.
