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Mathematics Question on Complex Numbers and Quadratic Equations

If α,β\alpha,\beta are the roots of the equation ax2+bx+c=0ax^2+bx+c = 0 and Sn=αn+βn,S_n = \alpha^n + \beta^n, then a Sn+1+bSn+cSn−1S_{n+1} + bS_{n} + cS_{n-1} is equal to

A

00

B

abcabc

C

a+b+ca + b +c

D

None of theseNone\, of\, these

Answer

00

Explanation

Solution

The correct option is(A): 0.

Given, α\alpha and β\beta are the roots of equation
ax2+bx+c=0ax^{2}+bx+c=0
∴α+β=−ba\therefore \alpha+\beta=-\frac{b}{a} and αβ=ca\alpha\beta =\frac{c}{a}
Now, Sn+1=αn+1+βn+1S_{n+1}=\alpha^{n+1}+\beta^{n+1}
=αn+1+βn+1+αnβ+βnα−αnβ−βnα=\alpha^{n+1}+\beta^{n+1}+\alpha^{n}\beta+\beta^{n}\alpha-\alpha^{n}\beta-\beta^{n}\alpha
=αn(α+β)+βn(α+β)−αβ(αn−1+βn−1)=\alpha^{n}\left(\alpha+\beta\right)+\beta^{n}\left(\alpha+\beta\right)-\alpha\beta \left(\alpha^{n-1}+\beta^{n-1}\right)
=(α+β)(αn+βn)−αβ(αn−1+βn−1)=\left(\alpha+\beta\right)\left(\alpha^{n}+\beta^{n}\right)-\alpha\beta\left(\alpha^{n-1}+\beta^{n-1}\right)
=(α+β)(αn+βn)−αβ(αn−1+βn−1)=\left(\alpha+\beta\right)\left(\alpha^{n}+\beta^{n}\right)-\alpha\beta \left(\alpha^{n-1}+\beta^{n-1}\right)
=−baSn−caSn−1=-\frac{b}{a} S_{n}-\frac{c}{a}S_{n-1}
⇒Sn+1=−bSn−cSn−1a\Rightarrow S_{n+1}=\frac{-bS_{n}-cS_{n-1}}{a}
∴aSn+1+bSn+cSn−1=0\therefore aS_{n+1}+bS_{n}+cS_{n-1}=0