Q118 / 200 · WBJEE 2014 · Mathematics

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

  • Correct answer
    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
    $S_{n}+S_{n-1}=2 S_{n+1}$

Solution

Given, $\alpha$ and $\beta$ are the roots of $x^{2}-x-1=0$ $\therefore \quad \alpha+\beta=1$ and $\alpha \beta=-1$ Now, consider option (a). $S_{n}+S_{n-1}=S_{n+1}$ Put $n=2$, we get, $\quad S_{2}+S_{1}=S_{3}$ $\therefore \quad\left(\alpha^{2}+\beta^{2}\right)+(\alpha+\beta)=\alpha^{3}+\beta^{3}$ $\left[\because \cdot S_{n}=\alpha^{n}+\beta^{n},\right.$ given] $\Rightarrow(a+\beta)^{2}-2 a \beta+(\alpha+\beta)$ $=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$ $\Rightarrow 1^{2}-2(-1)+1=1^{3}-3(-1)(1)$ $\Rightarrow 2+1=1+3$ $\Rightarrow \quad 1+2+1=1+3$ $\Rightarrow$ $4=4,$ which is true. Hence, option (a) is correct.

Hence, the correct answer is option: 'A'.

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