Question

Let $\alpha ,\beta$ be the roots of the equation $a{x}^2+bx+c=0,a\ne 0$, then sum of the series $(\alpha +\beta {)}^2+({\alpha }^2+{\beta }^2)+(\alpha –\beta {)}^2+........$ upto $n$ terms is equal to

• A. $\frac{n(b^2+(n-1\left)ac\right)}{a^2}$
• B. $\frac{n(b^2-(n-1\left)ac\right)}{a^2}$
• C. $\frac{n(a^2-(n-1\left)bc\right)}{b^2}$
• D. $\frac{n(a^2+(n-1\left)ac\right)}{a^2}$

Answer 1

The correct option is B $\frac{n(b^2-(n-1\left)ac\right)}{a^2}$

$\alpha +\beta =–b/a,\alpha \beta =c/a$
$(\alpha +\beta {)}^2+({\alpha }^2+{\beta }^2)+(\alpha -\beta {)}^2+..........$ upto $n$ terms
= $\frac{n}{2}[2{(\alpha +\beta )}^2+(n-1\left)\right(-2\alpha \beta \left)\right]$
= $\frac{n}{2}[2{(-b/a)}^2+(n-1\left)\right(-2c/a\left)\right]$
= $\frac{n}{2}[\frac{2b^2}{a^2}-2(n-1\left)\frac{c}{a}\right]$
=$\frac{n[b^2-(n-1\left)ac\right]}{a^2}$