JEE Main & Advanced Mathematics Sequence & Series Question Bank Self Evaluation Test - Sequences and Series

  • question_answer
    \[{{\left( x+\frac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}+{{\left( {{x}^{{}}}+\frac{1}{{{x}^{3}}} \right)}^{2}}....\] upto n terms is

    A) \[\frac{{{x}^{2n}}-1}{{{x}^{2}}-1}\times \frac{{{x}^{2n+2}}+1}{{{x}^{2n}}}+2n\]

    B) \[\frac{{{x}^{2n}}+1}{{{x}^{2}}+1}\times \frac{{{x}^{2n+2}}-1}{{{x}^{2n}}}-2n\]

    C) \[\frac{{{x}^{2n}}-1}{{{x}^{2}}-1}\times \frac{{{x}^{2n}}-1}{{{x}^{2n}}}-2n\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] The series is \[({{x}^{2}}+{{x}^{4}}+{{x}^{6}}\,\,up\,\,to\,\,n\,\,terms)\] \[+\left( \frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{4}}}+\frac{1}{{{x}^{6}}}+\,\,\text{up}\,\,\text{to}\,\,\text{n}\,\,\text{terms} \right)\text{+2+}\,\,\text{up}\,\,\text{to}\,\,\text{n}\,\,\text{terms)}\] \[a=\frac{{{x}^{2}}({{x}^{2n}}-1)}{{{x}^{2}}-1}+\frac{\frac{1}{{{x}^{2}}}\left( 1-\frac{1}{{{x}^{2n}}} \right)}{1-\frac{1}{{{x}^{2}}}}+2n\] \[=\frac{{{x}^{2}}({{x}^{2n}}-1)}{{{x}^{2}}-1}+\frac{{{x}^{2n}}-1}{({{x}^{2}}-1){{x}^{2n}}}+2n\] \[=\,\,\,\frac{{{x}^{2n}}-1}{{{x}^{2}}-1}\times \frac{{{x}^{2n+2}}+1}{{{x}^{2n}}}+2n\]


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