Question

Let \(a\) and \(b\) be the roots of the quadratic equation \(x^2 + 3x - 1 = 0\). If \(P_n = a^n + b^n\) for \(n \geq 0\), then for \(n \geq 2\), \(P_n =\)

• A. \(-3P_{n-1}+P_{n-2}\)
• B. \(3P_{n-1}+P_{n-2}\)
• C. \(-P_{n-1}+3P_{n-2}\)
• D. \(P_{n-1}+3P_{n-2}\)

Hint:

Since a and b satisfy x^2 = 1 - 3x, multiply through by x^(n-2) to link Pn with P(n-1) and P(n-2).

Answer 1

The Correct Option is A

Step 1: Write down what a and b satisfy.
Since a and b are the two roots of \(x^2 + 3x - 1 = 0\), both numbers satisfy this same equation. Rearranging it gives \(x^2 = 1 - 3x\), true for \(x = a\) and for \(x = b\).

Step 2: Multiply by \(x^{n-2}\) to bring in higher powers.
Take the equation \(x^2 = 1 - 3x\) and multiply both sides by \(x^{n-2}\), valid for any \(n \geq 2\):
\[ x^n = x^{n-2} - 3x^{n-1} \]
This holds separately for \(x = a\) and \(x = b\), since both satisfy the original quadratic.

Step 3: Add the two versions together.
For \(x = a\): \(a^n = a^{n-2} - 3a^{n-1}\).
For \(x = b\): \(b^n = b^{n-2} - 3b^{n-1}\).
Adding these two equations:
\[ a^n + b^n = (a^{n-2} + b^{n-2}) - 3(a^{n-1} + b^{n-1}) \]

Step 4: Rewrite in terms of Pn.
By definition \(P_n = a^n + b^n\), so the left side is \(P_n\), and the right side is \(P_{n-2} - 3P_{n-1}\). This gives:
\[ P_n = -3P_{n-1} + P_{n-2} \]

Step 5: Rule out the other options.
Options B and D flip the sign of the \(P_{n-1}\) term, which would come from the wrong equation \(x^2 = 3x - 1\), not the one actually given. Option C swaps which term carries the factor of 3, again mismatching the multiplication step above. Only the relation from Step 4 matches the quadratic in the question.

Final Answer:
\[ \boxed{P_n = -3P_{n-1} + P_{n-2}} \]