Question

If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0,a\ne 0$ and $S_n={\alpha }^n+{\beta }^n,$ then

• A. $a{S}_{n+1}+b{S}_n+c{S}_{n-1}=0$
• B. $S_5=-\frac{b}{a^5}(b^2-2ac{)}^2-\frac{(b^2-ac)bc}{a^4}$
• C. $a{S}_{n+1}+b{S}_n+c{S}_{n-1}=1$
• D. $S_5=-\frac{b}{a^5}(b^2-2ac{)}^2+\frac{(b^2-ac)bc}{a^4}$

Answer 1

Answer: (A), (D)

$S_5=-\frac{b}{a^5}(b^2-2ac{)}^2+\frac{(b^2-ac)bc}{a^4}$

We know that,
$\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0,a\ne 0,$
$a{\alpha }^2+b\alpha +c=0$
$a{\beta }^2+b\beta +c=0$
$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$

Now,
$S_n={\alpha }^n+{\beta }^n$
$a{S}_{n+1}+b{S}_n+c{S}_{n-1}$
$=a({\alpha }^{n+1}+{\beta }^{n+1})+b({\alpha }^n+{\beta }^n)+c({\alpha }^{n-1}+{\beta }^{n-1})$
$={\alpha }^{n-1}(a{\alpha }^2+b\alpha +c)+{\beta }^{n-1}(a{\beta }^2+b\beta +c)$
$={\alpha }^{n-1}\times 0+{\beta }^{n-1}\times 0$
$=0$

$\therefore S_{n+1}=-\frac{b}{a}{S}_n-\frac{c}{a}{S}_{n-1}...\left(1\right)$
Putting $n=4$ in the above equation, we get
$S_5=-\frac{b}{a}{S}_4-\frac{c}{a}{S}_3$
Now,
$S_4={\alpha }^4+{\beta }^4=\left(\right(\alpha +\beta {)}^2-2\alpha \beta {)}^2-2{\alpha }^2{\beta }^2={(\frac{b^2}{a^2}-\frac{2c}{a})}^2-\frac{2c^2}{a^2}=\frac{(b^2-2ac{)}^2}{a^4}-\frac{2c^2}{a^2}$
$S_3={\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=-\frac{b^3}{a^3}+\frac{3bc}{a^2}$
Putting in the expression of $S_5$,
$S_5=-\frac{b}{a^5}(b^2-2ac{)}^2+\frac{2b{c}^2}{a^3}+\frac{b^{3}c}{a^4}-\frac{3b{c}^2}{a^3}\Rightarrow S_5=-\frac{b}{a^5}(b^2-2ac{)}^2+\frac{b^{3}c}{a^4}-\frac{b{c}^2}{a^3}$

$\therefore S_5=-\frac{b}{a^5}(b^2-2ac{)}^2+\frac{(b^2-ac)bc}{a^4}$