Let alpha and beta be the roots of x^2-x-1=0, | vedclass

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(A) Given $x^2-x-1=0$,roots are $\alpha, \beta$. Thus $\alpha+\beta=1$ and $\alpha\beta=-1$.$a_n = \frac{\alpha^n-\beta^n}{\alpha-\beta}$.For $(1)$: $

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$1, 2, 4$

Vedclass · Answer

(A) Given $x^2-x-1=0$,roots are $\alpha, \beta$. Thus $\alpha+\beta=1$ and $\alpha\beta=-1$.$a_n = \frac{\alpha^n-\beta^n}{\alpha-\beta}$.For $(1)$: $a_{n+2}-a_{n+1} = \frac{\alpha^{n+2}-\beta^{n+2}-\alpha^{n+1}+\beta^{n+1}}{\alpha-\beta} = \frac{\alpha^{n+1}(\alpha-1)-\beta^{n+1}(\beta-1)}{\alpha-\beta}$. Since $\alpha^2-\alpha-1=0 \implies \alpha^2-1=\alpha$,we have $a_{n+2}-a_{n+1} = \frac{\alpha^{n+1}(\alpha^2-1)-\beta^{n+1}(\beta^2-1)}{\alpha-\beta}$ is incorrect logic; rather $a_{n+2} = a_{n+1}+a_n$. Thus $\sum_{i=1}^n a_i = a_{n+2}-a_2 = a_{n+2}-1$. Statement $(1)$ is correct.For $(2)$: $\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{1}{\alpha-\beta} \left( \sum (\frac{\alpha}{10})^n - \sum (\frac{\beta}{10})^n \right) = \frac{1}{\alpha-\beta} \left( \frac{\alpha/10}{1-\alpha/10} - \frac{\beta/10}{1-\beta/10} \right) = \frac{1}{10-(\alpha+\beta)+(\alpha\beta/10)} = \frac{10}{100-10-1} = \frac{10}{89}$. Statement $(2)$ is correct.For $(4)$: $b_n = a_{n-1}+a_{n+1} = \frac{\alpha^{n-1}-\beta^{n-1}+\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta} = \frac{\alpha^{n-1}(1+\alpha^2)-\beta^{n-1}(1+\beta^2)}{\alpha-\beta}$. Since $1+\alpha^2 = \alpha^2-\alpha\beta = \alpha(\alpha-\beta)$,we get $b_n = \alpha^n+\beta^n$. Statement $(4)$ is correct.Thus,options $1, 2, 4$ are correct.

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