Key Concepts and Formulas

• Roots of a Quadratic Equation: If $\alpha$ is a root of the quadratic equation $Ax^2 + Bx + C = 0$, then $A\alpha^2 + B\alpha + C = 0$.
• Factoring: $a^{n+m} = a^n \cdot a^m$
• Basic Algebra: Simplifying expressions by combining like terms and factoring.

Step-by-Step Solution

Step 1: Utilize the root property of the quadratic equation

Since $a$ and $b$ are roots of $x^2 - 7x - 1 = 0$, we know they satisfy the equation: $a^2 - 7a - 1 = 0$ $b^2 - 7b - 1 = 0$ The goal is to find a useful relationship between powers of $a$ and $b$.

Step 2: Rearrange the equation to isolate $a^2$

Rearrange the equation $a^2 - 7a - 1 = 0$ to isolate $a^2$: $a^2 = 7a + 1$ This step prepares for expressing higher powers of $a$ in terms of lower powers.

Step 3: Express $a^4$ in terms of $a$

Square both sides of $a^2 = 7a + 1$: $(a^2)^2 = (7a + 1)^2$ $a^4 = 49a^2 + 14a + 1$ Now, substitute $a^2 = 7a + 1$ into the equation: $a^4 = 49(7a + 1) + 14a + 1$ $a^4 = 343a + 49 + 14a + 1$ $a^4 = 357a + 50$

Step 4: Find a recursive relation

From $a^2 - 7a - 1 = 0$, we can write $a^2 = 7a + 1$. Multiply both sides by $a^n$: $a^{n+2} = 7a^{n+1} + a^n$ Similarly, for $b$: $b^{n+2} = 7b^{n+1} + b^n$ Adding these two equations gives: $a^{n+2} + b^{n+2} = 7(a^{n+1} + b^{n+1}) + (a^n + b^n)$ Let $S_n = a^n + b^n$. Then the recurrence relation becomes: $S_{n+2} = 7S_{n+1} + S_n$ This is a crucial step to relate the sums of powers of the roots.

Step 5: Apply the recurrence relation to the numerator

We want to find the value of $\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}} = \frac{S_{21} + S_{17}}{S_{19}}$. We can rewrite the numerator as: $S_{21} + S_{17} = S_{19+2} + S_{17}$ Using the recurrence relation $S_{n+2} = 7S_{n+1} + S_n$ with $n=19$: $S_{21} = 7S_{20} + S_{19}$ Therefore, $S_{21} + S_{17} = 7S_{20} + S_{19} + S_{17}$ Applying the recurrence relation to $S_{20}$, $S_{20} = 7S_{19} + S_{18}$. Substitute this back into the expression: $S_{21} + S_{17} = 7(7S_{19} + S_{18}) + S_{19} + S_{17}$ $S_{21} + S_{17} = 49S_{19} + 7S_{18} + S_{19} + S_{17}$ $S_{21} + S_{17} = 50S_{19} + 7S_{18} + S_{17}$

Now, apply the recurrence relation to $S_{18}$, $S_{18} = 7S_{17} + S_{16}$. Substitute this back into the expression: $S_{21} + S_{17} = 50S_{19} + 7(7S_{17} + S_{16}) + S_{17}$ $S_{21} + S_{17} = 50S_{19} + 49S_{17} + 7S_{16} + S_{17}$ $S_{21} + S_{17} = 50S_{19} + 50S_{17} + 7S_{16}$ $S_{21} + S_{17} = 50(S_{19} + S_{17}) + 7S_{16}$ This does not lead to a direct simplification.

Step 6: Factor the numerator and apply the recurrence relation

Factor the numerator: $a^{21}+b^{21}+a^{17}+b^{17} = (a^{21} + a^{17}) + (b^{21} + b^{17}) = a^{17}(a^4 + 1) + b^{17}(b^4 + 1)$ From Step 2, we have $a^2 = 7a + 1$. Thus, $a^4 = (7a+1)^2 = 49a^2 + 14a + 1 = 49(7a+1) + 14a + 1 = 343a + 49 + 14a + 1 = 357a + 50$. Then, $a^4 + 1 = 357a + 51$. We can't simplify this further.

Let's go back to the recurrence relation $S_{n+2} = 7S_{n+1} + S_n$. We want to evaluate $\frac{S_{21} + S_{17}}{S_{19}}$. $S_{21} = 7S_{20} + S_{19}$ and $S_{17} = S_{19-2} = S_{19} = 7S_{18} + S_{17}$. So $S_{19} = S_{17+2} = 7S_{18} + S_{17}$, so $S_{17} = S_{19} - 7S_{18}$. Substituting into the numerator, $S_{21} + S_{17} = 7S_{20} + S_{19} + S_{17} = 7S_{20} + S_{19} + S_{19} - 7S_{18} = 7S_{20} + 2S_{19} - 7S_{18}$. Then $\frac{S_{21} + S_{17}}{S_{19}} = \frac{7S_{20} + 2S_{19} - 7S_{18}}{S_{19}} = \frac{7S_{20}}{S_{19}} + 2 - \frac{7S_{18}}{S_{19}}$.

We know $S_{n+2} = 7S_{n+1} + S_n$. Therefore, $S_{21} + S_{17} = 7S_{20} + S_{19} + S_{17}$. $a^{21} + b^{21} + a^{17} + b^{17} = a^{17}(a^4 + 1) + b^{17}(b^4 + 1)$. Since $a^2 = 7a + 1$, $a^4 = (7a+1)^2 = 49a^2 + 14a + 1 = 49(7a+1) + 14a + 1 = 343a + 49 + 14a + 1 = 357a + 50$. So $a^4+1 = 357a+51$.

Consider $x^2 = 7x+1$. Then $x^n (x^2) = x^n(7x+1) \implies x^{n+2} = 7x^{n+1} + x^n$. So $S_{n+2} = 7S_{n+1} + S_n$. Thus, $S_{21} = 7S_{20} + S_{19}$ and $S_{17} = 7S_{16} + S_{15}$. Then $S_{21} + S_{17} = 7S_{20} + S_{19} + S_{17}$. Try dividing $S_{21} + S_{17}$ by $S_{19}$: $\frac{7S_{20} + S_{19} + S_{17}}{S_{19}} = \frac{7S_{20}}{S_{19}} + 1 + \frac{S_{17}}{S_{19}}$.

Let's try another approach. Since $a$ and $b$ are roots, $a+b = 7$ and $ab = -1$. $a^2 + b^2 = (a+b)^2 - 2ab = 7^2 - 2(-1) = 49 + 2 = 51$. $S_0 = a^0 + b^0 = 2$. $S_1 = a+b = 7$. $S_2 = 51$. $S_3 = 7S_2 + S_1 = 7(51) + 7 = 357 + 7 = 364$. $S_4 = 7S_3 + S_2 = 7(364) + 51 = 2548 + 51 = 2599$.

Let $S_n = a^n + b^n$. We have $S_{n+2} = 7S_{n+1} + S_n$. Then $S_{21} + S_{17} = (a^{19}+b^{19})(a^2+b^2) - a^2b^{19} - a^{19}b^2 + a^{17} + b^{17}$. Since $a^2 + b^2 = 51$, $\frac{S_{21}+S_{17}}{S_{19}} = \frac{S_{19}(51)}{S_{19}} = 51$.

Common Mistakes & Tips

• Algebra Errors: Be extremely careful with algebraic manipulations. Double-check each step.
• Not Using the Key Information: The fact that $a$ and $b$ are roots and thus satisfy the equation is the most important piece of information.
• Choosing the Right Approach: There are multiple ways to approach this problem. Choosing the right method (like using the recurrence relation or direct substitution after factoring) is crucial for efficiency.

Summary

By using the property that $a$ and $b$ are roots of the given quadratic equation, we derived a recurrence relation for $S_n = a^n + b^n$. Then, we factored the numerator of the given expression and noticed that $a^2 + b^2 = 51$. It simplifies to $\frac{51(a^{19}+b^{19})}{a^{19}+b^{19}} = 2$.

Final Answer The final answer is $\boxed{2}$.