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(D) Given the equation $x^{2}+5 \sqrt{2} x+10=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $\alpha^{2} + 5 \sqrt{2} \alpha + 10

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$1$

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(D) Given the equation $x^{2}+5 \sqrt{2} x+10=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $\alpha^{2} + 5 \sqrt{2} \alpha + 10 = 0 \implies \alpha^{2} = -5 \sqrt{2} \alpha - 10$ and $\beta^{2} = -5 \sqrt{2} \beta - 10$.Also,$P_{n} = \alpha^{n} - \beta^{n}$.Consider the expression $E = \frac{P_{17} P_{20} + 5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19} + 5 \sqrt{2} P_{18}^{2}}$.Factor out $P_{17}$ in the numerator and $P_{18}$ in the denominator:$E = \frac{P_{17}(P_{20} + 5 \sqrt{2} P_{19})}{P_{18}(P_{19} + 5 \sqrt{2} P_{18})}$.Substitute $P_{n} = \alpha^{n} - \beta^{n}$:$P_{20} + 5 \sqrt{2} P_{19} = (\alpha^{20} - \beta^{20}) + 5 \sqrt{2} (\alpha^{19} - \beta^{19}) = \alpha^{19}(\alpha + 5 \sqrt{2}) - \beta^{19}(\beta + 5 \sqrt{2})$.From the quadratic equation,$\alpha + 5 \sqrt{2} = -\frac{10}{\alpha}$ and $\beta + 5 \sqrt{2} = -\frac{10}{\beta}$.Substituting these:$P_{20} + 5 \sqrt{2} P_{19} = \alpha^{19}(-\frac{10}{\alpha}) - \beta^{19}(-\frac{10}{\beta}) = -10 \alpha^{18} + 10 \beta^{18} = -10 P_{18}$.Similarly,$P_{19} + 5 \sqrt{2} P_{18} = \alpha^{18}(\alpha + 5 \sqrt{2}) - \beta^{18}(\beta + 5 \sqrt{2}) = -10 \alpha^{17} + 10 \beta^{17} = -10 P_{17}$.Thus,$E = \frac{P_{17}(-10 P_{18})}{P_{18}(-10 P_{17})} = 1$.

If $\alpha, \beta$ are roots of the equation $x^{2}+5 \sqrt{2} x+10=0$,$\alpha > \beta$ and $P_{n}=\alpha^{n}-\beta^{n}$ for each positive integer $n$,then the value of $\left(\frac{P_{17} P_{20}+5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}\right)$ is equal to $....$

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