Question

Let $\alpha ,\beta$ be the roots of $x^2-x-1=0$ and $S_n={\alpha }^n+{\beta }^n$, for all integers $n\ge 1$. Then for every integer $n\ge 2$

• A. $S_n+S_{n-1}=S_{n+1}$
• B. $S_n-S_{n-1}=S_{n+1}$
• C. $S_{n-1}=S_{n+1}$
• D. $S_n+S_{n-1}=2S_{n+1}$

Answer 1

Answer: (A)

$S_n+S_{n-1}=S_{n+1}$

The roots of the equation $x^2-x-1=0$ are $\alpha$ and $\beta$

So the roots will satisiy the equation $x^2-x-1=0$

$\Rightarrow$ $(\alpha {)}^2-(\alpha )-1=0$, $(\beta {)}^2-(\beta )-1=0$

Now from above equation we can get

$(\alpha {)}^2=(\alpha )+1$........eq (1)

$(\beta {)}^2=(\beta )+1$ .........eq (2)

Now multiply equation $\left(1\right)$ by $(\alpha {)}^{n-1}$ and equation $\left(2\right)$ by $(\beta {)}^{n-1}$, we get

$(\alpha {)}^{n+1}=(\alpha {)}^n+(\alpha {)}^{n-1}$ ..... $\left(3\right)$

$(\beta {)}^{n+1}=(\beta {)}^n+(\beta {)}^{n-1}$ ....... $\left(4\right)$

Now adding $\left(3\right)$ and $\left(4\right)$, we get

$(\beta {)}^{n+1}+(\alpha {)}^{n+1}=(\alpha {)}^{n-1}+(\beta {)}^{n-1}+(\alpha {)}^n+(\beta {)}^n$ and according to question we can write it as

$S_{n+1}=S_n+S_{n-1}$

Hence, option A is correct.