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 $$\frac{P_{17}P_{20} + 5\sqrt{2}P_{17}P_{19}}{P_{18}P_{19} + 5\sqrt{2}P_{18}^2}$$ is equal to ___.

We start from the quadratic equation $$x^{2}+5\sqrt{2}\,x+10=0$$ whose roots are given to be $$\alpha$$ and $$\beta$$ with $$\alpha >\beta$$. From the standard relation between coefficients and roots of a quadratic, we immediately obtain

$$\alpha+\beta=-5\sqrt{2},\qquad \alpha\beta=10.$$

For every positive integer $$n$$ we define $$P_{n}=\alpha^{n}-\beta^{n}$$ and wish to evaluate

$$E=\dfrac{P_{17}P_{20}+5\sqrt{2}\,P_{17}P_{19}}{P_{18}P_{19}+5\sqrt{2}\,P_{18}^{2}}.$$

Because both $$\alpha$$ and $$\beta$$ satisfy the quadratic, we may replace $$x$$ by either root in the equation $$x^{2}+5\sqrt{2}\,x+10=0$$. Writing the result for an arbitrary positive power will give us a recurrence for the sequence $$P_{n}$$.

First, for any root $$r\in\{\alpha,\beta\}$$ we have $$r^{2}=-5\sqrt{2}\,r-10.$$ Multiplying both sides by $$r^{\,n-2}$$ (where $$n\ge 2$$) gives

$$r^{n}= -5\sqrt{2}\,r^{\,n-1}-10\,r^{\,n-2}.$$

Since this equality holds separately for $$r=\alpha$$ and for $$r=\beta$$, we can write the two identities and subtract them. Thus, for all integers $$n\ge 2$$,

$$\alpha^{n}-\beta^{n}= -5\sqrt{2}\left(\alpha^{\,n-1}-\beta^{\,n-1}\right)-10\left(\alpha^{\,n-2}-\beta^{\,n-2}\right).$$

Recognizing the differences as the $$P$$ sequence, we arrive at the linear recurrence relation

$$P_{n}=-5\sqrt{2}\,P_{n-1}-10\,P_{n-2}\qquad (n\ge 2).$$

Now we tackle the pieces that appear in the expression $$E$$.

First, apply the recurrence with $$n=20$$:

$$P_{20}=-5\sqrt{2}\,P_{19}-10\,P_{18}.$$

Add $$5\sqrt{2}\,P_{19}$$ to both sides to create the combination seen in the numerator:

$$P_{20}+5\sqrt{2}\,P_{19}=(-5\sqrt{2}\,P_{19}-10\,P_{18})+5\sqrt{2}\,P_{19}=-10\,P_{18}.$$

Hence the entire numerator becomes

$$P_{17}\bigl(P_{20}+5\sqrt{2}\,P_{19}\bigr)=P_{17}\times(-10\,P_{18})=-10\,P_{17}P_{18}.$$

Next, apply the recurrence with $$n=19$$ to prepare the denominator:

$$P_{19}=-5\sqrt{2}\,P_{18}-10\,P_{17}.$$

Add $$5\sqrt{2}\,P_{18}$$ to both sides:

$$P_{19}+5\sqrt{2}\,P_{18}=(-5\sqrt{2}\,P_{18}-10\,P_{17})+5\sqrt{2}\,P_{18}=-10\,P_{17}.$$

Hence the entire denominator is

$$P_{18}\bigl(P_{19}+5\sqrt{2}\,P_{18}\bigr)=P_{18}\times(-10\,P_{17})=-10\,P_{17}P_{18}.$$

Now both numerator and denominator are equal to the same quantity $$-10\,P_{17}P_{18}$$, so when we form their ratio we get

$$E=\dfrac{-10\,P_{17}P_{18}}{-10\,P_{17}P_{18}}=1.$$

So, the answer is $$1$$.

Hence, the correct answer is Option 1.