Lucas number identity Let $L_n$ be the Lucas numbers, defined by the recursion $L_n=L_{n-1}+L_{n-2}$ with initial values $L_0=2$ and $L_1=1$.
Any idea how to prove the identity 
$$\sum_{j\ge{0}}(-1)^{n-j}\left(\binom{2n}{n+j+1}+\binom{2n+2}{n+j+2}\right)(L_{2j+2}-2)=[n=0]$$
or$$\sum_{j\ge{0}}(-1)^{n-j}\binom{2n+2}{n+j+2} (L_{2j+2}-2)=1?$$
 A: Here is an answer using the Binet formula for the Lucas number:
$$L_j=\phi^j+(-\phi^{-1})^j$$
where $\tag{1} \ \ \ \phi=\frac{1+\sqrt5}{2}, \ \  1-\phi^2=-\phi \  \ \text{    and    }\ \ 1-\phi^{-2}=\phi^{-1}.$
Let $ S_n(x)=\sum_{j=0}^n(-1)^j {2n\choose{n+j}}x^j$, then we have
$$\begin{align} S_n(x)&=\frac{1}{2}\left(\sum_{j=0}^n(-1)^j {2n\choose{n+j}}x^j+\sum_{j=0}^n(-1)^j {2n\choose{n+j}}x^j\right)\\
&=\frac{1}{2}\left(\sum_{j=0}^n(-1)^j {2n\choose{n+j}}x^j+\sum_{j=0}^n(-1)^j {2n\choose{n-j}}x^j\right)\\
&=\frac{1}{2}\left(\sum_{j=n}^{2n}(-1)^{j-n} {2n\choose{j}}x^{j-n}+\sum_{j=0}^n(-1)^{n-j} {2n\choose{j}}x^{n-j}\right)\\
S_n(x^{-1})&=\frac{1}{2}\left(\sum_{j=n}^{2n}(-1)^{j-n} {2n\choose{j}}x^{n-j}+\sum_{j=0}^n(-1)^{n-j} {2n\choose{j}}x^{j-n}\right)\\
\end{align}$$
Then
\begin{align}
\sum_{j=0}^{n}(-1)^j {2n\choose{n+j}}\left(x^j+x^{-j}\right)&=\frac{1}{2}\left(\sum_{j=0}^{2n}(-1)^{j-n} {2n\choose{j}}x^{n-j}+\sum_{j=0}^{2n}(-1)^{n-j} {2n\choose{j}}x^{j-n} \right)\\
&+{2n \choose n}
\end{align}
then
$$\tag2 \sum_{j=0}^{n}(-1)^j {2n\choose{n+j}}\left(x^j+x^{-j}\right)={2n \choose n}+\frac{(-1)^n}{2}\left(x^n(1-x^{-1})^{2n}+x^{-n}(1-x)^{2n}\right)$$
Now we specialize $(2)$ with $x=1$  so that 
$$\tag3 {2n \choose n}=2\sum_{j=0}^{n}(-1)^j {2n\choose{n+j}}-[n=0]$$
and with  $x=\phi^2$ so that,  making use of (1) and (3)
$$\begin{align}
\sum_{j=0}^{n}(-1)^j {2n\choose{n+j}}L_{2j}&={2n \choose n}+\frac{(-1)^n}{2}\left(\phi^{2n}(1-\phi^{-2})^{2n}+\phi^{-2n}(1-\phi^2)^{2n}\right)\\
&={2n \choose n}+ (-1)^n  \\
&=2\sum_{j=0}^{n}(-1)^j {2n\choose{n+j}}-[n=0]+ (-1)^n.
\end{align}
$$
then writing the same equation for $n+1$ instead of $n$, adding the two and accounting for $[n+1=0]=0$, we arrive at 
$$\begin{align}\sum_{j=0}^{n}(-1)^j \left({2n\choose{n+j}}+{2n+2\choose{n+1+j}}\right)L_{2j}&=2\sum_{j=0}^{n}(-1)^j \left({2n\choose{n+j}}+{2n+2\choose{n+1+j}}\right)\\
&-[n=0]\end{align}$$
which is the OP identity with a shift in the dum index $j$.
