Prove that after rearranging the above equation, we get
$$\begin{aligned} (3.2 .8) \Longleftrightarrow & 3 f_{0.4}^{(9)}+12 f_{0.5}^{(9)}+(-8+2 k) f_{1,1}^{(9)}+(-20+8 k) f_{1,2}^{(9)}+(1+14 k) f_{1,3}^{(9)} \\ & +(56+34 k) f_{1,4}^{(9)}+\left(-48+28 k+2 k^{2}\right) f_{2,1}^{(9)}+\left(8+8 k+5 k^{2}\right) f_{2,2}^{(9)} \end{aligned}$$

When $k \geq 4$, we have
$$2 k-8 \geq 0, \quad 8 k-20 \geq 0, \quad 1+14 k \geq 0, \quad 56+34 k \geq 0, \quad 2 k^{2}+28 k-48 \geq 0, \quad 5 k^{2}+8 k+8 \geq 0$$

Thus, we know that (3.2.8) holds. Proof completed.