Evaluating $sumlimits_{i=lceil frac{n}{2}rceil}^inftybinom{2i}{n}frac{1}{2^i}$

First, as Steven Stadnicki's question comment states, you can also get the same result as what you're asking for by starting "from $i = 0$" because "$binom{a}{b} = 0$ when $a lt b$" (note $binom{a}{b}$ means the number of ways to choose $b$ items from among $a$ items, but there are $0$ ways to do this if $b gt a$).

Next, for positive natural numbers $n$ and $k$, Pascal's rule states

$$binom{n}{k} = binom{n - 1}{k} + binom{n - 1}{k - 1} tag{1}label{eq1A}$$

With positive natural numbers $i$ and $n$, this gives

$$binom{2i}{n + 1} = binom{2i - 1}{n + 1} + binom{2i - 1}{n} tag{2}label{eq2A}$$

Using eqref{eq1A} on both terms of the RHS of eqref{eq2A} gives

$$binom{2i - 1}{n + 1} = binom{2i - 2}{n + 1} + binom{2i - 2}{n} tag{3}label{eq3A}$$

$$binom{2i - 1}{n} = binom{2i - 2}{n} + binom{2i - 2}{n - 1} tag{4}label{eq4A}$$

Substituting these into eqref{eq2A}, multiplying both sides by $frac{1}{2^{i}}$, summing from $i = 1$ to $infty$, changing the summation indices on the RHS and making a few algebraic manipulations, you end up with

$$begin{equation}begin{aligned} binom{2i}{n + 1} & = binom{2i - 2}{n + 1} + 2binom{2i - 2}{n} + binom{2i - 2}{n - 1} \ binom{2i}{n + 1}frac{1}{2^i} & = frac{1}{2}binom{2i - 2}{n + 1}frac{1}{2^{i-1}} + binom{2i - 2}{n}frac{1}{2^{i-1}} + frac{1}{2}binom{2i - 2}{n - 1}frac{1}{2^{i-1}} \ sum_{i=1}^{infty}binom{2i}{n + 1}frac{1}{2^i} & = frac{1}{2}sum_{i=1}^{infty}binom{2(i - 1)}{n + 1}frac{1}{2^{i-1}} + sum_{i=1}^{infty}binom{2(i - 1)}{n}frac{1}{2^{i-1}} + frac{1}{2}sum_{i=1}^{infty}binom{2(i - 1)}{n - 1}frac{1}{2^{i-1}} \ a_{n+1} & = frac{1}{2}sum_{i=0}^{infty}binom{2i}{n + 1}frac{1}{2^{i}} + sum_{i=0}^{infty}binom{2i}{n}frac{1}{2^{i}} + frac{1}{2}sum_{i=0}^{infty}binom{2i}{n - 1}frac{1}{2^{i}} \ a_{n+1} & = left(frac{1}{2}right)a_{n+1} + a_{n} + left(frac{1}{2}right)a_{n-1} \ left(frac{1}{2}right)a_{n+1} & = a_{n} + left(frac{1}{2}right)a_{n-1} \ a_{n+1} & = 2a_n + a_{n-1} end{aligned}end{equation}$$