What is an induction step in proving, that the following system of equations has these solutions: $d_i={\binom{2i-1}{i}}^2\dfrac{1}{4^{2i-1}}$
\begin{align} \binom31=&\binom21+4d_1\binom00\\ \binom52+\binom50d_1=&\binom42+4d_1\binom21+4^2d_2\binom00\\ \binom73+\binom71d_1=&\binom63+4d_1\binom42+4^2d_2\binom21+4^3d_3\binom00\\ \binom94+\binom92d_1+\binom90d_2=&\binom84+4d_1\binom63+4^2d_2\binom42+4^3d_3\binom21+4^4d_4\binom00\\ \binom{11}5+\binom{11}3d_1+\binom{11}1d_2=&\binom{10}5+4d_1\binom84+4^2d_2\binom63+4^3d_3\binom42+4^4d_4\binom21+4^5d_5\binom00\\ \vdots \end{align}
$$\sum_{i=0}^{\left \lfloor{\frac n2}\right \rfloor }\binom{2n+1}{n-2i}d_i=\sum_{i=0}^n4^id_i\binom{2n-2i}{n-i} \quad n=1,2,3,\ldots$$