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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$$\sum_{j\ge{0}}(-1)^{n-j}\binom{2n+2}{n+j+2} (L_{2j+2}-2)=1?$$

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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 $j$.

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  • $\begingroup$ Thank you for this very nice proof! $\endgroup$ – Johann Cigler Jan 22 at 9:45

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