Question

Let $\alpha$ and $\beta$ be the roots of the equation, $5x^2+6x-2=0$. If $S_n={\alpha }^n+{\beta }^n$, $n=1,2,3,..$ then

• A. $5S_6+6S_5+2S_4=0$
• B. $6S_6+5S_5=2S_4$
• C. $6S_6+5S_5+2S_4=0$
• D. $5S_6+6S_5=2S_4$

Answer 1

The correct option is D

$5S_6+6S_5=2S_4$

Explanation for the correct option:

The given quadratic equation:$5x^2+6x-2=0$

And $\alpha$ and $\beta$ be the roots of this equation

Then substituting $x=\alpha$ in this equation

$$\begin{gathered} \Rightarrow 5{\alpha }^2+6\alpha -2=0 \\ \Rightarrow 6\alpha -2=-5{\alpha }^2..\left(i\right) \end{gathered}$$

Similarly, by substituting $x=\beta$ we have,

$\Rightarrow 6\beta -2=-5{\beta }^2..\left(ii\right)$

Since given $S_n={\alpha }^n+{\beta }^n$

Therefore,

$$\begin{gathered} S_6={\alpha }^6+{\beta }^6...\left(iii\right) \\ {S}_5={\alpha }^5+{\beta }^5...\left(iv\right) \\ {S}_4={\alpha }^4+{\beta }^4...\left(v\right) \end{gathered}$$

Now by,

$6\times \left(iv\right)-2\times \left(v\right)$ we have

$$\begin{gathered} \Rightarrow 6S_5-2S_4=6{\alpha }^5+6{\beta }^5-2{\alpha }^4-2{\beta }^4 \\ ={\alpha }^4(6\alpha -2)+{\beta }^4(6\beta -2) \\ ={\alpha }^4(-5{\alpha }^2)+{\beta }^4(-5{\beta }^2) \\ =-5({\alpha }^6+{\beta }^6) \\ =-5S_6 \end{gathered}$$

Thus, $6S_5+5S_6=2S_4$

Hence, the correct option is (D)