Question

If $ \alpha,\beta $ are the roots of the equation $ ax^2+bx+c = 0 $ and $ S_n = \alpha^n + \beta^n, $ then a $ S_{n+1} + bS_{n} + cS_{n-1} $ is equal to

• A. $0$
• B. $abc$
• C. $a + b +c$
• D. $None\, of\, these$

Answer 1

The Correct Option is A

The correct option is(A): 0.

Given, \(\alpha\) and \(\beta\) are the roots of equation 
\(ax^{2}+bx+c=0\)
\(\therefore \alpha+\beta=-\frac{b}{a}\) and \(\alpha\beta =\frac{c}{a}\)
Now, \(S_{n+1}=\alpha^{n+1}+\beta^{n+1}\)
\(=\alpha^{n+1}+\beta^{n+1}+\alpha^{n}\beta+\beta^{n}\alpha-\alpha^{n}\beta-\beta^{n}\alpha\)
\(=\alpha^{n}\left(\alpha+\beta\right)+\beta^{n}\left(\alpha+\beta\right)-\alpha\beta \left(\alpha^{n-1}+\beta^{n-1}\right)\)
\(=\left(\alpha+\beta\right)\left(\alpha^{n}+\beta^{n}\right)-\alpha\beta\left(\alpha^{n-1}+\beta^{n-1}\right)\)
\(=\left(\alpha+\beta\right)\left(\alpha^{n}+\beta^{n}\right)-\alpha\beta \left(\alpha^{n-1}+\beta^{n-1}\right)\)
\(=-\frac{b}{a} S_{n}-\frac{c}{a}S_{n-1}\)
\(\Rightarrow S_{n+1}=\frac{-bS_{n}-cS_{n-1}}{a}\)
\(\therefore aS_{n+1}+bS_{n}+cS_{n-1}=0\)