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] \[\sum{\frac{{{a}^{n}}}{{{10}^{n}}}=\frac{1}{\alpha -\beta }\left\{ \frac{\alpha }{10-\alpha }-\frac{\beta }{10-\beta } \right\}}\]\[=\,\frac{1}{\alpha -\beta }\left\{ \frac{10(\alpha -\beta )}{89} \right\}\,=\,\frac{10}{89}.\] |
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