Q53 / 125 · BITSAT 2016 · Mathematics

$\left(x+\frac{1}{x}\right)^{2}+\left(x^{2}+\frac{1}{x^{2}}\right)^{2}+$ $\left(x^{3}+\frac{1}{x^{3}}\right)^{2} \quad \ldots .$ upto $n$ terms is

  • Correct answer
    A
    $\frac{x^{2 n}-1}{x^{2}-1} \times \frac{x^{2 n+2}+1}{x^{2 n}}+2 n$
  • B
    $\frac{x^{2 n}+1}{x^{2}+1} \times \frac{x^{2 n+2}-1}{x^{2 n}}-2 n$
  • C
    $\frac{x^{2 n}-1}{x^{2}-1} \times \frac{x^{2 n}-1}{x^{2 n}}-2 n$
  • D
    None of these

Solution

The series is $$ \begin{array}{l} \left(x^{2}+x^{4}+x^{6}+\ldots\right)+\left(\frac{1}{x^{2}}+\frac{1}{x^{4}}+\frac{1}{x^{6}}+\ldots\right) \\ +(2+2+\ldots . .) \\ =\frac{x^{2}\left(x^{2 n}-1\right)}{x^{2}-1}+\frac{\frac{1}{x^{2}}\left(1-\frac{1}{x^{2 n}}\right)}{1-\frac{1}{x^{2}}}+2 n \\ =\frac{x^{2}\left(x^{2 n}-1\right)}{x^{2}-1}+\frac{x^{2 n}-1}{\left(x^{2}-1\right) x^{2 n}}+2 n \\ =\frac{x^{2 n}-1}{x^{2}-1} \times \frac{x^{2 n+2}+1}{x^{2 n}}+2 n \end{array} $$

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

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