• # question_answer Let $\alpha$ and $\beta$ be the roots of ${{x}^{2}}-x-1=0,$ with $\alpha \,>\,\beta .$ For all positive integer n, define ${{a}_{n}}\,=\,\frac{{{\alpha }^{n}}\,-\,{{\beta }^{n}}}{\alpha \,-\,\beta },\,n\ge \,1$, ${{b}_{1}}\,=\,1$ and ${{b}_{n}}\,=\,{{a}_{n-1}}\,+\,{{a}_{n}}+{{1}^{,\,}}\,n\,\,\le \,2$. Then which of the following options is/are correct? A) ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{n}}={{a}_{n+2}}-\,1$ for all $n\,\,\ge \,\,1$ B) ${{b}_{n}}={{a}^{n}}+{{\beta }^{n}}$ for all $n\,\ge \,1$ C) $\sum\limits_{n\,=\,1}^{\infty }{\frac{{{b}_{n}}}{{{10}^{n}}}\,=\,\frac{8}{89}}$ D) $\sum\limits_{n\,=\,1}^{\infty }{\frac{{{a}_{n}}}{{{10}^{n}}}\,=\,\frac{10}{89}}$

Correct Answer: A , B , D

Solution :

 ${{x}^{2}}-x-1=0$ ${{a}_{n}}=\frac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta }$ [b]        ${{b}_{1}}=1$ ${{b}_{n}}=\,{{a}_{n-1}}+{{a}_{n+1}}$ $\alpha =\frac{1+\sqrt{5}}{2},\beta =\frac{1-\sqrt{5}}{2}$ ${{b}_{n}}=\frac{{{\alpha }^{n\,-\,1}}-{{\beta }^{n\,-\,1}}}{\alpha -\beta }+\frac{{{\alpha }^{n+1}}+{{\beta }^{n+1}}}{\alpha -\beta }$$=\frac{{{\alpha }^{n-1}}(1+{{\alpha }^{2}})-{{\beta }^{n-1}}(1+{{\beta }^{2}})}{\alpha -\beta }$$=\,\frac{{{\alpha }^{n\,-\,1}}\left( \frac{5+\sqrt{5}}{2} \right)-{{\beta }^{n\,-\,1}}\left( \frac{5-\sqrt{5}}{2} \right)}{\alpha -\beta }$ $=\,\frac{\sqrt{5}{{\alpha }^{n}}+\sqrt{5}{{\beta }^{n}}}{\alpha -\beta }={{\alpha }^{n}}+{{\beta }^{n}}$ (i)         ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}$$=\,\frac{(\alpha +{{\alpha }^{2}}+...+{{\alpha }^{n}})-(\beta +{{\beta }^{2}}+...{{\beta }^{n}})}{\alpha -\beta }$$=\,\frac{\frac{\alpha (1-{{\alpha }^{n}})}{1-\alpha }-\frac{\beta (1-{{\beta }^{n}})}{1-\beta }}{\alpha -\beta }$ ${{\alpha }^{2}}-\alpha -1=0$ ${{\alpha }^{2}}-1=\alpha$ $\alpha +1=\frac{\alpha }{\alpha -1}$$=\,\frac{-{{\alpha }^{2}}(1-{{\alpha }^{n}})+{{\beta }^{2}}(1-{{\beta }^{n}})}{\alpha -\beta }$$=\frac{-{{\alpha }^{2}}+{{\alpha }^{n+2}}+{{\beta }^{2}}-{{\beta }^{n+2}}}{(\alpha -\beta )}$$=\,\frac{{{\alpha }^{n+2}}-{{\beta }^{n+2}}}{(\alpha -\beta )}-\,(\alpha +\beta )$$={{\alpha }_{n+2}}-1.$ [c]         $\sum{\frac{{{b}_{n}}}{{{10}^{n}}}=\sum{\left( \frac{{{\alpha }^{n}}}{{{10}^{n}}}+\frac{{{\beta }^{n}}}{{{10}^{n}}} \right)}}$         $=\,\left( \frac{\alpha }{10}+\frac{{{\alpha }^{2}}}{{{10}^{2}}}+... \right)$$=\,\frac{\frac{\alpha }{10}}{1\,-\,\frac{\alpha }{10}}+\frac{\beta }{1\,-\,\frac{\beta }{10}}$$=\,\frac{\alpha }{10-\alpha }+\frac{\beta }{10-\beta }$$=\,\frac{10(\alpha +\beta )-2\alpha \beta }{100-10(\alpha +\beta )+\alpha \beta }$ $=\,\frac{10+2}{100-10-1}=\,\frac{12}{89}$ [d]        $\sum{\frac{{{a}^{n}}}{{{10}^{n}}}=\frac{1}{\alpha -\beta }\left\{ \frac{\alpha }{10-\alpha }-\frac{\beta }{10-\beta } \right\}}$$=\,\frac{1}{\alpha -\beta }\left\{ \frac{10(\alpha -\beta )}{89} \right\}\,=\,\frac{10}{89}.$

Solution :

Same as above

Solution :

Same as above

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