$\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$ is a convex function on $[0,+\infty)$? Let $f(x)=\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$, where $u\in\mathbb{Z}^+$ and $\binom{x}{l}=\frac{x(x-1)\dots(x-l+1)}{l!}$ for all $l\in\mathbb{Z}^+$. 
Then can we prove $f(x)$ is a convex function on $[0,+\infty)$?
Updates:
1) It was pointed out by @user44191 that, observing $\binom{x}{i}=\binom{x-1}{i}+\binom{x-1}{i-1}$, the question is equivalent to $\binom{x-1}{1}+\binom{x-1}{2}+\dots+\binom{x-1}{2u}$ is convex on $[0,+\infty)$.
2) Pointed out by @FedorPetrov @GeraldEdgar @H.H.Rugh:
For $x<0$ each summand $\binom{x}{2i}$ is obviously convex, thus the question is equivalent to $f(x)$ is convex on $\mathbb{R}$.
3) Pointed out by @WłodzimierzHolsztyń‌​ski:
It has $(\Delta^2 f_u)(x) =  1+ f_{u-1}(x-2)$, where $(\Delta f_u)(x)=f_u(x)-f_u(x-1)$. Then we can conclude that $f(x)$ is discrete convex.
 A: Here's an alternative proof based on probabilistic arguments (showing different aspects). Let 
$$f_n(x):=\sum_{j=0}^n   { x  \choose j}=[t^n]\,\frac{(1+t)^x}{1-t}\;\;,$$
and let $^\prime$ denote derivative with resp. to $x$.
We have to show that for even $n=2k$ the second derivative $f_{2k}^{\prime\prime}(x)=[t^{2k}]
\frac{(1+t)^x}{1-t}\,(\log(1+t))^2$ is nonnegative.
The basic observation used below is that $g(t):=\frac{\log(1+t)}{t}$ is the Laplace transform (LT) of a nonnegative random variable  possessing all moments. 
The relation $g(t)=\int_0^1 \frac{1}{1+st}\,ds$ shows that $g$ is the LT of $U\cdot X_1$,  where $U$ is uniform on $[0,1]$, $X_1$ is $\Gamma(1,1)=\exp(1)$ distributed
(i.e. has LT  $\frac{1}{1+t}$), and the factors 
are independent.
Case 1: 
$x<0$. In this case $(1+t)^x $ is the LT of  $\Gamma(1, -x)$ . Thus $\ell(t):=(1+t)^x g(t)^2$ is the Laplace transform of a nonnegative rv 
and (for $k\geq 1$)  $f_{2k}^{\prime\prime}(x)=[t^{2k-2}]\, \frac{1}{1-t} \ell(t)$ is an even degree MacLaurin sum of $\ell$, evaluated  at $1$, 
and therefore exceeds $\ell(1)>0$.
Case 2:
$x\geq 0$. Since $f_n(x+1)=f_n(x)+f_{n-1}(x)$ it will suffice to show that $f_n^{\prime\prime}(x)\geq 0$ for $x\in [0,1)$ and all $n$.
For $x=0$ all derivatives $f_n^{\prime\prime}(0)$ are nonnegative, since $a_r:=[t^r] (g(t))^2=2\,(-1)^r \frac{H_{r+1}}{r+2}$ , 
and $a_0=1>0$, $a_{2k}+a_{2k+1}\geq 0$.
Let $0<x<1$ and write $$(1+t)^x =1 +xt\,\frac{ (1+t)^x -1}{xt} =1 +xt\,h_x(t)$$ and accordingly
$$f_n^{\prime\prime}(x)=[t^{n-2}] \frac{1}{1-t}\left(1+ x t\,h_x(t)\right)g(t)^2=f_n^{\prime\prime}(0)+ x [t^{n-3}]\frac{h_x(t)g(t)^2}{1-t}\;\;.$$
Here $h_x$ is the LT of $U\cdot X_{1-x}$, where $U$ is uniform on $[0,1]$, $X_{1-x}$ is $\Gamma(1,1-x)$ (LT $(1+t)^{x-1}$) distributed, and the factors 
are independent, thus $h_x(t)g(t)^2$ is again the LT of a random variable $Z$ posessing all moments.
(1) If $n$ is odd, $n-3$ is even and $f_n^{\prime\prime}(x)\geq f^{\prime\prime}(0)$, since the second term is nonnnegative by the same argument as above.
(2) If $n$ is even, $n-3$ is odd and the second term is the $Z$-expection of a decreasing function, and will therefore not increase if $Z$ is replaced by a stochastically 
larger random variable. Replacing $U\cdot X_1$ for $U\cdot X_{1-x}$ in the first factor replaces $g$ for $h_x$, makes $Z$ stochastically larger and we get
 $$f_n^{\prime\prime}(x)\geq f_n^{\prime\prime}(0)+ x [t^{n-3}]\frac{g(t)^3}{1-t}=f_n^{\prime\prime}(0)+x\,f_n^{\prime\prime\prime}(0)\;\;.$$
If $f_n^{\prime\prime\prime}(0)\geq 0$ we're done. If  $f_n^{\prime\prime\prime}(0) <0$ it will suffice to show that 
$f_n^{\prime\prime}(0)+f_n^{\prime\prime\prime}(0)\geq 0$. 
This amounts to showing that the (even) partial sums of the coefficients 
of $c(t):=(\log(1+t))^2 + (\log(1+t))^3 $ are nonnegative, and this can be done
EDIT: some details:  the $n$-th coefficient $c_n=[t^n] c(t)$ of $c$ is $$c_n=\frac{(-1)^n}{n}\left(2 H_{n-1}-3(H_{n-1}^2-H_{n-1}^{(2)})\right)$$
Hence
$$c_n+c_{n+1}=\frac{(-1)^n}{n(n+1)}\left(-3 H_{n-1}^2 + 8 H_{n-1} + 3 H_{n-1}^{(2)} -2\right)$$
Thus (for $2k\geq 2$) the even partial sums fall until $2k=n+1=12$  and rise thereafter. Since $[t^{12}]\frac{c(t)}{1-t}=\frac{26647}{221760}>0.12$, all even partial sums are nonnegative.
Finally, note that this proof also shows that $f_n$ is convex on $[0,\infty)$ for odd $n$.
A: (Here is a proof of the convexity of $\sum_{k=0}^n\binom{x}{k}$ on $[0,\infty)$ 
for any large $n$; with a bit more care the argument should work for all positive even integer $n$ on $[-1,\infty)$, which is the original problem)
We may consider  the problem of showing the convexity of  $S_n(x):=\sum_{k=0}^n\binom{x}{k}$ on $[-1,+\infty)$ for any positive even integer $n$, which is equivalent to the original, as observed. As also observed, since $S_n$ is smooth and each summand is convex for $x>n-1$, it is sufficient to prove the convexity on $[-1,n-1]$. For odd $n$,  the convexity should be true only on $[0,+\infty)$ .
The sum $S_n$ is the $n$-th Taylor polynomial of the function $(1+t)^x$, centered at $t_0=0$ and evaluated at $t=1$. The corresponding remainder integral formula gives:
$$\sum_{k=0}^n\binom{x}{k}= 2^{x} - (n+1)\binom{x}{n+1}\int_0^1(1-t)^{n}(1+t)^{-n-1+x}dt.$$
So in order to prove convexity of the right-hand side on some interval $I=I_n$ we need  to show the inequality
$$2^x(\log 2)^2 - \int_0^1(n+1)(1-t)^{n}(1+t)^{-n-1}  \bigg[(1+t)^x\binom{x}{n+1}\bigg]''dt\ge0,$$
for $x\in I$ (here $'$ denotes derivative wrto $x$). Note that the integral weight $(n+1)(1-t)^{n}(1+t)^{-n-1}$ in front of the second derivative has mass less than $1$ (for a quick check: up to a sign, $ \int_0^1(1-t)^{n}(1+t)^{-n-1} dt$ is again an integral remainder of a Taylor expansion, namely of  $\log(1+t)$, and in fact its value is exactly the remainder of the logarithmic series $\big|\log(2)-\sum_{k=1}^n(-1)^k/k\big|$, which is not larger that $1/(n+1)$.)
Therefore
$$\int_0^1(n+1)(1-t)^{n}(1+t)^{-n-1}  \bigg[(1+t)^x\binom{x}{n+1}\bigg]''dt \le\sup_{0\le t\le 1} \bigg[(1+t)^x\binom{x}{n+1}\bigg]''   $$
$$=\sup_{0\le t\le 1} (1+t)^x\bigg[ \log^2(1+t)\binom{x}{n+1} +2\log(1+t) \binom{x}{n+1}'+\binom{x}{n+1}'' \bigg] $$
$$\le 2^x(1+\log 2)^2 \max\bigg\{\binom{x}{n+1}, \binom{x}{n+1}', \binom{x}{n+1}'' \bigg\}. $$
We can conclude that for a given positive integer $n$, $S_n(x)$ is convex for $x\in I$, provided the uniform norms of $\binom{x}{n+1}$, and of its first and second derivative   on the interval $I$ is uniformly less than $1/6$.
Since these uniform norms on $I=[0,n]$ converge to $0$ as $n\to\infty$, it follows the convexity of $S_n(x)$ on $[0,\infty)$ for any large $n$. 
(I do not have handy the relative convergence bounds, that should be classic in polynomial interpolation. For even $n$, by experiments it seems they are less than $1/6$ as soon as $n\ge8$).   
Also note that the above weight concentrates around $t=0$, which suggests to break  the integral into the two intervals $[0,\tau]$ and $[\tau, 1]$, to be estimated separately; this gives a much better estimate, of course, and should be of use in the original problem for even $n$, which required the bound for $x\in[-1,0]$ too.
A: This is not an answer to your question, is only an equivalent reformulation that seems promising. I write it as an answer only because of space constraints.
For any nonnegative integer $n$  and any  $\newcommand{\bR}{\mathbb{R}}$ $x\in\bR$ we define
$$ a_n(x)=\sum_{k=0}^n \binom{x}{2k}, \;\;b_n(x)=\sum_{k=0}^n \binom{x}{2k+1},$$
where for any  nonnegative integer $m$ we set
$$
\binom{x}{m}:=\frac{x(x-1)\cdots (x-m+1)}{m!}.
$$
For $t\in (-1,1)$  and  $x\in\bR$ the series
$$ F_x(t):=\sum_{k=0}^\infty \binom{x}{k}t^k $$
is the Taylor series at $t=0$ of the function $t\mapsto (1+t)^x$ and  thus
$$ F_x(t)=(1+t)^x,\;\;\forall |t|<1. $$
The generating series of the even binomial coefficients $\binom{x}{2m}$ is then
$$ F^0_x(t)=\sum_{m\geq 0}\binom{x}{2m}t^{2m}= \frac{1}{2}\Bigl(\, F_x(t)+F_x(-t)\,\Bigr)=\frac{(1+t)^x+(1-t)^x}{2}. $$
The  generating series of the odd binomial coefficients $\binom{x}{2m+1}$ is then
$$ F^1_x(t)=\sum_{m\geq 0}\binom{x}{2m+1}t^{2m}= \frac{1}{2}\Bigl(\, F_x(t)-F_x(-t)\,\Bigr)=\frac{(1+t)^x-(1-t)^x}{2}. $$
The generating series  of the  sequence
$$a_n(x)=\sum_{k=0}^n \binom{x}{2k} $$
is
$$ A_x(t)=\sum_{n\geq 0} a_n(x)t^{2n}=\frac{1}{1-t^2} F_x^0(t)=\frac{(1+t)^x+(1-t)^{x}}{2(1-t^2)}. $$
The generating series of $b_n(x)$ is
$$
B_x(t)=\sum_{n\geq 0} b_n(x)t^{2n+1}=\frac{1}{1-t^2} F_x^1(t)=\frac{(1+t)^x-(1-t)^{x}}{2(1-t^2)}. $$
We have $\newcommand{\pa}{\partial}$
$$\pa^2_x A_x(t)=\sum_{n\geq 0}  a_n''(x) t^{2n}. $$
The problem  is equivalent  to showing that, for any $x\in \bR$,  the Taylor coefficients at $t=0$ of the function
$$[0,1)\ni t\mapsto \pa^2_xA_x(t) $$
are nonnegative, i.e. for any $x$, the function $t\mapsto \pa^2_x A_x(t)$ is absolutely monotonic on the $t$-interval $[0,1)$; for definition and properties of absolutely monotonic functions see Chap. IV of  Widder's classical monograph The Laplace Transform.
Now observe that
$$ \pa^2_xA_x(t)=\frac{(1+t)^x\bigl(\,\log(1+t)\,\bigr)^2+(1-t)^x\bigl(\,\log(1-t)\,\bigr)^2}{2(1-t^2)}. $$
Actually we only need to prove that the even degree Taylor coefficients  at $t=0$ of  the function
$$ t\mapsto  G_x(t)=(1+t)^x\frac{\log^2(1+t)}{1-t^2} $$
are positive for any $x\in\bR$.
Remark. As observed in comments to the question,  we can instead study the convexity of the function
$$ c_n(x)=\sum_{j=0}^{2n}\binom{x}{j}.$$
The generating series
$$ C_x(t) =\sum_{n\geq 0} c_n(x) t^{2n}, $$
is the even part of
$$ G_x(t)=\frac{(1+t)^x}{1-t}, $$
i.e.,
$$ C_x(t)=\frac{1}{2}\left(  \frac{(1+t)^x}{1-t}+ \frac{(1-t)^x}{1+t}\right)=\frac{(1+t)^{x+1}+(1-t)^{x+1}}{2(1-t^2)} =A_{x+1}(t). $$
