Proving equations with binomial coefficients 
Let $n$ be a positive integer and let $x$ be a non zero real number. Prove the following.
  
  
*
  
*$\sum_{k=0}^n \dbinom{n}{k}2^{n-k} (1+x)(x^{-1}+x)^k = \frac{1}{x^n} (1+x)^{2n+1}$
  
*$\sum_{k=0}^n \dbinom{n}{k}2^{n-k} \dbinom{k}{\left\lfloor\frac{k}{2}\right\rfloor}= \dbinom{2n+1}{n}$
  

Sorry that I cannot make the title more clear. It is exactly my problem: I don't even know what the problem is asking, or say, what knowledge do I need to solve the problem.
It is a homework problem so I would appreciate some helps but not a complete solution. I have learnt to give combinatorial proof for some simple equalities, Sperner's Theorem, multinomial theorem and Newton's binomial theorem. But I just have no clue where to start and what to use. Thanks in advance.
 A: $[1.]$ just follows from the binomial theorem:
$$\sum_{k=0}^{n}\binom{n}{k}A^{n-k}B^k = (A+B)^n\tag{1} $$
and the fact that $(x^{-1}+x+2)=\frac{(x+1)^2}{x}$.
$[2.]$ can be solved by writing $\binom{2k}{k}$ as the coefficient of $x^k$ in $(1+x)^{2k}$. Assume $n=2m$.
$$ \sum_{k=0}^{m}\binom{2m}{2k}2^{2m-2k}\binom{2k}{k}=\sum_{k=0}^{m}\binom{2m}{2k}4^{m-k}\cdot[x^k](1+x)^{2k}\tag{2} $$
$$ \sum_{k=0}^{m-1}\binom{2m}{2k+1}2^{2m-2k-1}\binom{2k+1}{k}=\sum_{k=0}^{m-1}\binom{2m}{2k}2^{2m-2k-1}\cdot[x^k](1+x)^{2k+1}\tag{3} $$
and:
$$\sum_{k=0}^{m}\binom{2m}{2k}4^{m-k}B^k = \frac{1}{2}\left((2+\sqrt{B})^{2m}+(\sqrt{B}-2)^{2m}\right)\tag{4}$$
hence $(2)$ is the coefficient of $x^m$ in $\frac{1}{2}\left((x+3)^{2m}+(x-1)^{2m}\right)$, i.e. $\frac{3^m+(-1)^m}{2}\binom{2m}{m}$.
Can you finish now, by computing $(3)$ in the same way and studying the case $n=2m+1$, too?
As a faster alternative, multiply both sides of your $[1.]$ by $x^n$ and compute the coefficient of $x^n$ in the new RHS and LHS.
A: Here is a proof that leaves some  work for you to do in completing the
details.
Suppose we seek to verify that
$$\sum_{k=0}^n {n\choose k} 2^{n-k} {k\choose \lfloor k/2 \rfloor}
= {2n+1\choose n}.$$

This is
$$\sum_{q=0}^n {n\choose 2q} 2^{n-2q} {2q\choose q}
+ \sum_{q=0}^n {n\choose 2q+1} 2^{n-2q-1} {2q+1\choose q}.$$
We treat these in turn.
 First sum.
Observe that
$${n\choose 2q} {2q\choose q}
= {n\choose q} {n-q\choose q}.$$
This yields for the sum
$$2^n \sum_{q=0}^n {n\choose q} {n-q\choose q} 2^{-2q}.$$
Introduce
$${n-q\choose q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-q}}{z^{q+1}} \; dz$$
which yields for the sum
$$\frac{2^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z} 
\sum_{q=0}^n {n\choose q} 2^{-2q} \frac{1}{z^q(1+z)^q}
\; dz
\\ = \frac{2^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z} 
\left(1+\frac{1}{4z(1+z)}\right)^n
\; dz
\\ = \frac{2^{-n}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+2z)^{2n}}{z^{n+1}} \; dz
= 2^{-n} {2n\choose n} 2^n = {2n\choose n}.$$
 Second sum.
Observe that
$${n\choose 2q+1} {2q+1\choose q}
= {n\choose q} {n-q\choose q+1}.$$
This yields for the sum
$$2^{n-1} \sum_{q=0}^n {n\choose q} {n-q\choose q+1} 2^{-2q}.$$
This time introduce
$${n-q\choose q+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-q}}{z^{q+2}} \; dz$$
which yields for the sum
$$\frac{2^{n-1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^2} 
\sum_{q=0}^n {n\choose q} 2^{-2q} \frac{1}{z^q(1+z)^q}
\; dz
\\ = \frac{2^{n-1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^2} 
\left(1+\frac{1}{4z(1+z)}\right)^n
\; dz
\\ = \frac{2^{-n-1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+2z)^{2n}}{z^{n+2}} \; dz
= 2^{-n-1} {2n\choose n+1} 2^{n+1} = {2n\choose n+1}.$$
Conclusion.
Collecting the two contributions we obtain
$${2n\choose n}+{2n\choose n+1} = {2n+1\choose n}$$
as claimed.
