Question

If $\alpha ,\beta ,\gamma$ are roots of equation $x^3-2x^2-1=0$ and $T_n={\alpha }^n+{\beta }^n+{\gamma }^n$, then value of $\frac{T_{11}-T_8}{T_{10}}$ is equal to

• A. 1
• B. 2
• C. -1
• D. 3

Answer 1

Answer: (B)

$x^3-2x^2-1=0$ or ${\alpha }^3-1=2{\alpha }^2$
$\therefore \frac{T_{11}-T_8}{T_{10}}=\frac{{\alpha }^{11}+{\beta }^{11}+{\gamma }^{11}-\left[{\alpha }^8+{\beta }^8+{\gamma }^8\right]}{{\alpha }^{10}+{\beta }^{10}+{\gamma }^{10}}$
$=\frac{{\alpha }^8\left({\alpha }^3-1\right)+{\beta }^8\left({\beta }^3-1\right)+{\gamma }^8\left({\gamma }^3-1\right)}{{\alpha }^{10}+{\beta }^{10}+{\gamma }^{10}}$
$=\frac{2\left({\alpha }^{10}+{\beta }^{10}+{\gamma }^{10}\right)}{{\alpha }^{10}+{\beta }^{10}+{\gamma }^{10}}=2$