Question

Let \(\alpha,\beta\) be the roots of the equation \(x^2-ax+b=0\) and \(A_n=\alpha^n+\beta^n\). Then \(A_{n+1}-aA_n+bA_{n-1}\) is equal to

• A. \(-a\)
• B. \(b\)
• C. 0
• D. \(a-b\)

Hint:

If \(\alpha,\beta\) are roots of quadratic, powers satisfy recurrence derived from \(\alpha^2=a\alpha-b\).

Answer 1

The Correct Option is C

Step 1: Use that \(\alpha,\beta\) satisfy the quadratic.
\[ \alpha^2-a\alpha+b=0 \Rightarrow \alpha^2=a\alpha-b \]
\[ \beta^2-a\beta+b=0 \Rightarrow \beta^2=a\beta-b \]
Step 2: Multiply first by \(\alpha^{n-1}\).
\[ \alpha^{n+1}=a\alpha^n-b\alpha^{n-1} \]
Similarly:
\[ \beta^{n+1}=a\beta^n-b\beta^{n-1} \]
Step 3: Add both equations.
\[ \alpha^{n+1}+\beta^{n+1}=a(\alpha^n+\beta^n)-b(\alpha^{n-1}+\beta^{n-1}) \]
\[ A_{n+1}=aA_n-bA_{n-1} \]
Step 4: Rearrange.
\[ A_{n+1}-aA_n+bA_{n-1}=0 \]
Final Answer:
\[ \boxed{0} \]