Earlier [ https://math.stackexchange.com/questions/4841673/finding-b-k-if-1x-x2n-sum-k-02n-b-k-xk ] it has been shown that $$B_k=\sum_{j=0}^{k} (-1)^{k-j} {n \choose j}{j \choose k-j}.......(1)$$ And $$B_{2n-k}=(-1)^{n-k}B_k......(2)$$ If $g(x)=(1+x-x^2)^n=\sum_{k=0}^{2n} B_k x^k.$ $$g(x)=(1-ax)^n (1-bx)^n=\left(\sum_{j=0}^{n} {n \choose j} (-ax)^j \right) \left(\sum_{j=0}^{n} {n \choose j} (-bx)^j \right) $$ Using the Cauchy product of two series, we can write $$g(x)=\sum_{k=0}^{n} \sum_{j=0}^{k} {n \choose j} (-ax)^j {n \choose k-j} (-bx)^{k-j}.$$ $$=\sum_{k=0}^n x^k \sum_{j=0}^{k} (-1)^j {n \choose j} {n \choose k-j} a^{2j-k}$$ Hence $$B_k=\sum_{j=0}^{k} (-1)^j {n \choose j} {n \choose k-j} \phi^{2j-k},........(3)$$ where $\phi=a$ is the golden ratio and (2) will hold again.
Here, the question is: Can one prove the identity of (1) and (3) directly?