Question

If $\alpha$ and $\beta$ are the roots of $x^{2}-ax+b=0$ and if $\alpha^{n}+\beta^{n}=V_{_n},$ then

• A. $V_{_{n+1}}= aV_{_n}+ bV_{_{n-1}}$
• B. $V_{_{n+1}}= aV_{_n}+ aV_{_{n-1}}$
• C. $V_{_{n+1}}= aV_{_n}- bV_{_{n-1}}$
• D. $V_{_{n+1}}= aV_{_n-1}- bV_{_{n}}$

Answer 1

The Correct Option is C

Multiplying $x^{2}-a x+b=0$ by $x^{n-1}$ $x^{n+1}=a x^{n}+b x^{n-1}=0\ldots$(i) $\alpha, \beta$ are roots of $x^{2}-a x+b=0$, therefore they will satisfy (i). Also, $\alpha^{n+1}-a \alpha^{n}+b \alpha^{n-1}=0 \ldots$(ii) and $\beta^{n+1}-a \beta^{n}+b \beta^{n-1}=0$ On adding Eqs. (ii) and (iii), we get $\left(\alpha^{n+1}+\beta^{n+1}\right)-a\left(\alpha^{n}+\beta^{n}\right) $ $+b\left(\alpha^{n-1}+\beta^{(n-1)}\right)=0 $ or $V_{n+1}-a V_{n}+b V_{n-1}=0$ $\Rightarrow V_{n+1}=a V_{n}-b V_{n-1}$ (given, $\left.\alpha^{n}+\beta^{n}=V_{n}\right)$