I will write a problem with an answer that apparently is wrong. My question would be what is wrong with this solution.

Define $$B(s)=\sum_{i=0}^s{{2\,s-i-1\choose s-1}\frac {i}{s}{5}^{i}{2}^{s-i}}$$ for integers $s>0$. The problem is to find $f(x)$ such that $$\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s}f(x)dx=B(s)\qquad\text{and}\qquad \int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s-1}f(x)dx=0.\tag{$*$}$$

To find $f(x)$ we can use Zeilberg's algorithm and obtain that $B(s)$ satisfy the following recurrence relation $$25\,B \left( s \right) -3\,B \left( s+1 \right) =10\,{\frac {{8}^{s} \Gamma \left( s+1/2 \right) }{\sqrt {\pi }\Gamma \left( s+2 \right) }}.$$

Now, we can write the right hand side as an integral $$10\,{\frac {{8}^{s} \Gamma \left( s+1/2 \right) }{\sqrt {\pi }\Gamma \left( s+2 \right) }}=\int_{-2\sqrt{2}}^{2\sqrt{2}}{\frac {5{x}^{2\,s}\sqrt {8-{x}^{2}}}{2\pi }}dx.\tag{$**$}$$

Now replacing (*) and (**) in the recurrence, we obtain $$25\,\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s}f(x)dx -3\,\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s+2}f(x)dx=\int_{-2\sqrt{2}}^{2\sqrt{2}}{\frac {5{x}^{2\,s}\sqrt {8-{x}^{2}}}{2\pi }}dx.$$

Taking derivatives and solving for $f(x)$, we obtain $$f(x)={\frac {5\sqrt {8-{x}^{2}}}{2\pi \, \left( 25-3\,{x}^{2} \right) }}.$$

It seems everything is correct to me. However, if we take $s=1$ $$\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2}f(x)dx=\frac{20}{9},$$ but $$B(1)=5.$$

In fact it is wrong for every $s$ I fix.

Would you be kind and give me some ideas on how to fix my solution to find $f(x)$?