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

This question is really about Newton's identities, the standard trick that links power sums of roots to the coefficients of the equation they satisfy.

For the quadratic $x^2 + 3x - 1 = 0$, the sum of roots is $a + b = -3$ and the product is $ab = -1$ (comparing with $x^2 - (a+b)x + ab = 0$).

For a quadratic with roots $a, b$, the power sums $P_n = a^n + b^n$ always follow the recurrence:

\[ P_n = (a+b)P_{n-1} - ab\,P_{n-2} \]

This comes from expanding $(a+b)(a^{n-1}+b^{n-1}) = a^n + b^n + ab(a^{n-2}+b^{n-2})$, which rearranges to $P_n = (a+b)P_{n-1} - ab\,P_{n-2}$.

Plugging in $a + b = -3$ and $ab = -1$:

\[ P_n = (-3)P_{n-1} - (-1)P_{n-2} = -3P_{n-1} + P_{n-2} \]

Checking each option:

• Option A, $-3P_{n-1}+P_{n-2}$, matches exactly what we derived.
• Option B, $3P_{n-1}+P_{n-2}$, would need $a+b=3$, which is wrong since $a+b=-3$.
• Option C, $-P_{n-1}+3P_{n-2}$, mixes up which coefficient goes with which term.
• Option D, $P_{n-1}+3P_{n-2}$, gets both the sign and the placement wrong.

So the correct recurrence is $P_n = -3P_{n-1} + P_{n-2}$.