Question

If $\alpha ,\beta$ are the roots of quadratic equation $6x^2-2x+1=0$ and $S_n={\alpha }^n+{\beta }^n$, then which of the following statement(s) is/are true ?

• A. $6S_{n+1}-2S_n+S_{n-1}=0$
• B. $6S_{n+1}-2S_n+S_{n-3}=0$
• C. $\frac{2S_6-6S_7}{S_5}=1$
• D. $\frac{2S_6-6S_7}{S_5}=3$

Answer 1

Answer: (A), (C)

The correct options are
A $6S_{n+1}-2S_n+S_{n-1}=0$
C $\frac{2S_6-6S_7}{S_5}=1$

Given equation $6x^2-2x+1=0$ has roots $\alpha ,\beta$
$\therefore 6{\alpha }^2-2\alpha +1=0\Rightarrow (6\alpha -2+\frac{1}{\alpha })=0\Rightarrow {\alpha }^n(6\alpha -2+\frac{1}{\alpha })=0\Rightarrow 6{\alpha }^{n+1}-2{\alpha }^n+{\alpha }^{n-1}=0\cdots \left(1\right)$

Similarly, with root $\beta$
$6{\beta }^{n+1}-2{\beta }^n+{\beta }^{n-1}=0\cdots \left(2\right)$

Adding equation$\left(1\right)$ and $\left(2\right)$,
$6({\alpha }^{n+1}+{\beta }^{n+1})-2({\alpha }^n+{\beta }^n)+({\alpha }^{n-1}+{\beta }^{n-1})=0\Rightarrow 6S_{n+1}-2S_n+S_{n-1}=0(\because S_n={\alpha }^n+{\beta }^n)$

Now, $S_{n-1}=2S_n-6S_{n+1}$
$\Rightarrow \frac{2S_n-6S_{n+1}}{S_{n-1}}=1\cdots \left(3\right)$
Putting the value of $n=6$ in the equation $\left(3\right)$, we get
$\frac{2S_6-6S_7}{S_5}=1$